Mahesh

Practicing the AP Inter 2nd Year Maths Study Material and AP Inter 2nd Year Maths Chapter 7 Integrals Solutions Exercise 7j will help students to clear their doubts quickly.

Inter 2nd Year Maths Integrals Solutions Exercise 7j

Integrals Exercise 7j Solutions

I. Evaluate the following definite integrals

Question 1.
\(\int_0^4|x-1|\) dx
Solution:
Let I = \(\int_0^4|x-1|\) dx
It can be seen that, (x – 1) ≤ 0 when 0 ≤ x ≤ 1 and (x – 1) ≥ 0 when 1 ≤ x ≤ 4
∴ I = \(\int_0^1-(x-1) d x+\int_1^4(x-1)\) dx [∵ \(\int_a^b f(x) d x=\int_b^c f(x) d x+\int_c^b f(x) d x\)]
= \(\int_0^1(1-x) d x+\int_1^4(x-1)\) dx
= \(\left[x-\frac{x^2}{2}\right]_0^1+\left[\frac{x^2}{2}-x\right]_1^4\)
= (1 – \(\frac{1}{2}\)) – 0 + (\(\frac{4^2}{2}\) – 4) – (\(\frac{1}{2}\) – 1)
= \(\frac{1}{2}\) + 4 + \(\frac{1}{2}\) = 5

Question 2.
\(\int_2^8|x-5|\) dx
Solution:
Let I = \(\int_2^8|x-5|\) dx
It can be seen that, (x – 5) ≤ 0 on [2, 5] and (x – 5) ≥ 0 on [5, 8]
[∵ \(\int_a^b f(x) d x=\int_b^c f(x) d x+\int_c^b f(x) d x\)]
∴ I = \(\int_2^5\) {-(x – 5)} dx + \(\int_5^8\)(x – 5) dx
= \(\left[5 x-\frac{x^2}{2}\right]_2^5+\left[\frac{x^2}{2}-5 x\right]_5^8\)
= (25 – \(\frac{25}{2}\)) – (10 – \(\frac{4}{2}\)) + (\(\frac{64}{2}\) – 40) – (\(\frac{25}{2}\) – 25)
= \(\frac{25}{2}\) – 8 – 8 + \(\frac{25}{2}\) = 25 – 16 = 9

Question 3.
\(\int_{-5}^5\)|x+2| dx
Solution:
Let I = \(\int_{-5}^5\)|x+2| dx
It can be seen that, (x + 2) ≤ 0 on [-5, -2] and (x + 2) ≥ 0 and [-2, 5]
∴ \(\int_{-5}^{-2}\)-(x + 2) dx + \(\int_{-2}^5\)(x + 2) dx
[∵ \(\int_a^b f(x) d x=\int_b^c f(x) d x+\int_c^b f(x) d x\)]
I = \(-\left[\frac{x^2}{2}+2 x\right]_{-5}^{-2}+\left[\frac{x^2}{2}+2 x\right]_{-2}^5\)
= \(-\left[\frac{(-2)^2}{2}+2(-2)-\frac{(-5)^2}{2}-2(-5)\right]+\left[\frac{(5)^2}{2}+2(5)-\frac{(-2)^2}{2}-2(-2)\right]\)
= -[2 – 4 – \(\frac{25}{2}\) + 10] + [\(\frac{25}{2}\) + 10 – 2 + 4]
= \(\frac{25}{2}\) – 8 + \(\frac{25}{2}\) + 12 = 29

Question 4.
\(\int_0^1 \)x(1 – x)ⁿ dx
Solution:
Let I = \(\int_0^1 \)(1 – x)ⁿ dx
I = \(\int_0^1\)(1 – x){1 – (1 – x)}ⁿ dx
[∵ \(\int_0^{\mathrm{a}} \mathrm{f}(\mathrm{x}) \mathrm{dx}=\int_0^{\mathrm{a}} \mathrm{f}(\mathrm{a}-\mathrm{x}) \mathrm{dx}\)]
= \(\int_0^1\)(1 – x)xⁿ dx = \(\int_0^1\)(xⁿ – xⁿ⁺¹dx
= \(\left[\frac{x^{n+1}}{n+1}-\frac{x^{n+2}}{n+2}\right]_0^1\) = \(\left[\frac{1}{n+1}-\frac{1}{n+2}\right]\) – 0
= \(\frac{(n+2)-(n+1)}{(n+1)(n+2)}\)
= \(\frac{1}{(n+1)(n+2)}\)

Question 5.
\(\int_0^2 x \sqrt{2-x}\) dx
Solution:

Question 6.
\(\int_0^{\frac{\pi}{2}} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}\) dx
Solution:

Question 7.
\(\int_0^{\frac{\pi}{2}} \frac{\sin ^{\frac{3}{2}} x}{\sin ^{\frac{3}{2}} x+\cos ^{\frac{3}{2}} x}\) dx
Solution:

Question 8.
\(\int_0^{\pi / 2} \frac{\cos ^5 x}{\sin ^5 x+\cos ^5 x}\) dx
Solution:
Let I = \(\int_0^a f(x) d x=\int_0^a f(a-x)\) dx ……….. (i)
I = \(\int_0^{\pi / 2} \frac{\cos ^5 x}{\sin ^5 x+\cos ^5 x}\) dx
[∵ \(\int_0^{\mathrm{a}} \mathrm{f}(\mathrm{x}) \mathrm{dx}=\int_0^{\mathrm{a}} \mathrm{f}(\mathrm{a}-\mathrm{x}) \mathrm{dx}\)]
= \(\int_0^{\pi / 2} \frac{\sin ^5 x d x}{\sin ^5 x+\cos ^5 x}\) …………..(2)
On adding (1) and (2), we get
2I = \(\int_0^{\pi / 2} \frac{\cos ^5 x+\sin ^5 x}{\cos ^5 x+\sin ^5 x}\) dx
= \(\int_0^{\pi / 2}\) 1 dx = \([x]_0^{\pi / 2}\)
= \(\frac{\pi}{2}\) – 0
⇒ I = \(\frac{\pi}{4}\)

Question 9.
\(\int_0^{\frac{\pi}{2}} \frac{\sin x-\cos x}{1+\sin x \cos x}\) dx
Solution:
Let I = \(\int_0^{\frac{\pi}{2}} \frac{\sin x-\cos x}{1+\sin x \cos x} \) dx ………(i)
= \(\int_0^{\frac{\pi}{2}} \frac{\sin \left(\frac{\pi}{2}-x\right)-\cos \left(\frac{\pi}{2}-x\right)}{1+\sin \left(\frac{\pi}{2}-x\right) \cos \left(\frac{\pi}{2}-x\right)}\) dx
[∵ \(\int_0^{\mathrm{a}} \mathrm{f}(\mathrm{x}) \mathrm{dx}=\int_0^{\mathrm{a}} \mathrm{f}(\mathrm{a}-\mathrm{x}) \mathrm{dx}\)]
= \(\int_0^{\frac{\pi}{2}} \frac{\cos x-\sin x}{1+\sin x \cos x}\) dx ……….. (ii)
[∵ sin(\([x]_0^{\pi / 2}\) – x) = cos x and cos(\([x]_0^{\pi / 2}\) – x) – sin x]
On adding (1) and (2), we get
⇒ 2I = \(\int_0^{\frac{\pi}{2}} \frac{0}{1+\sin x \cos x}\) dx
⇒ I = 0

Question 10.
\(\int_0^a \frac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}}\) dx
Solution:
Let I = \(\int_0^a \frac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}}\) ……(1)
= \(\frac{\sqrt{a-x}}{\sqrt{a-x}+\sqrt{a-(a-x)}}\) dx
(∵ \(\int_0^{\mathrm{a}} \mathrm{f}(\mathrm{x}) \mathrm{dx}=\int_0^{\mathrm{a}} \mathrm{f}(\mathrm{a}-\mathrm{x}) \mathrm{dx}\))
I = \(\int \frac{\sqrt{a-x}}{\sqrt{a-x}+\sqrt{x}}\) dx ……….(2)
Adding (1) and (2), we get
⇒ 2I = \(\int_0^a \frac{\sqrt{x}+\sqrt{a-x}}{\sqrt{x}+\sqrt{a-x}}\) dx
= \(\int_0^a\) 1 . dx = \([x]_0^a\)
= a – 0 = a ⇒ I = \(\frac{a}{2}\)

Question 11.
\(\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\) sin⁷ dx
Solution:
Let I = \(\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\) sin⁷ dx .
Here f(x) = sin⁷
f(-x) = sin⁷(-x) = [-sin x]⁷ = -sin⁷x
∴ f(-x) = -f(x)
So, f(x) is an odd function, then ⇒ \(\int_{\frac{\pi}{2}}^{\frac{\pi}{2}}\) sin⁷dx = 0
[∵ \(\int_{-a}^a\) f(x) dx = 0, if f(x) is an odd function]

Question 12.
\(\frac{\int_{-\pi}^{\frac{\pi}{2}}}{2}\)sin²x dx
Solution:
Let I = \(\frac{\int_{-\pi}^{\frac{\pi}{2}}}{2}\)sin²x dx
f(x) = sin²x
f(-x) = sin²(-x) = [sin (-x)]² = (-sin x)²
∴ f(x) is an even function
∴ I = \(\frac{\int_{-\pi}^{\frac{\pi}{2}}}{2}\)sin²x dx = 2\(\int_0^{\pi / 2}\)sin²x dx
[∵ \(\int_{-a}^a\) f(x) dx = 2\(\int_0^a\) f(x) dx, if f(x) is an even function]
= \(2 \int_0^{\pi / 2}\left[\frac{1-\cos 2 x}{2}\right]\)dx ∵ cos 2x = 1 – 2sin²x
= \(\int_0^{\pi / 2}\)(1 – cos 2x) dx = \(\left[x-\frac{\sin 2 x}{2}\right]_0^{\pi / 2}\)
= \(\left[\frac{\pi}{2}-\frac{\sin \pi}{2}\right]\) – (0 – 0) = \(\frac{\pi}{2}\) – 0 = \(\frac{\pi}{2}\)

Question 13.
\(\int_{-1}^1\)x¹⁷ cos⁴x dx
Solution:
Let I = \(\int_{-1}^1\)x¹⁷ cos⁴x dx
Put x = -t, then dx = -dt
L.L : If x = -1 ⇒ t = 1 & U.L : If x = 1 ⇒ t = -1
∴ I = \(\int_{-1}^1\)(-t)¹⁷ cos⁴(-t) -(dt)
= \(\int_{-1}^1\) -t¹⁷ cos⁴t (-dt) = \(\int_{-1}^1\)-t¹⁷ cos⁴(-t) (dt)
It is an odd function
∴ I = \(\int_{-1}^1\)x¹⁷ cos⁴x dx = 0

Question 14.
\(\int_0^{2 \pi} \) cos⁵ x dx
Solution:
Let I = \(\int_0^{2 \pi}\) cos⁵ x dx
Now cos⁵(2π – x) = cos⁵ x
∴ I = 2\(\int_0^\pi \)cos⁵ x dx
[∵ \(\int_0^{2 \mathrm{a}}\) f(x) dx = 2\(\int_0^{\mathrm{a}}\) f(x) dx, where f(2a – x) = f(x)]]
Now cos⁵(π – x) = – cos⁵x⁰ = 2(0) = 0
[∵ \(\int_0^{2 \mathrm{a}}\) f(x)dx = 0, where f(2a – x) = f(x)]

III. Evaluate the following definite integrals

Question 1.
\(\int_0^{\pi / 4}\) log (1 + tan x) dx
Solution:

Question 2.
\(\int_0^1 \frac{\log (1+x)}{1+x^2}\) dx
Solution:

Question 3.
\(\int_0^{\frac{\pi}{2}}(2 \log \sin x-\log \sin 2 x) d x\)
Solution:

Question 4.
\(\int_0^\pi \frac{x}{1+\sin x} d x\)
Solution:

Question 5.
\(\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sin x+\cos x}{\sqrt{\sin 2 x}}\) dx
Solution:
Let I = \(\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sin x+\cos x}{\sqrt{\sin 2 x}}\)dx
Put sin cos x = t (cos x + sin x) dx = dt
⇒ (sin x – cos x)² = t²
⇒ sin² x + cos² x – 2 sin x cos x = t²
⇒ 1 – sin 2x = t² ⇒ 1 – t² = sin 2x
L.L: x = π/6 then t = sin π/6 – cos π/6
= \(\frac{1}{2}-\frac{\sqrt{3}}{2}=\frac{1-\sqrt{3}}{2}\)

Question 6.
\(\int_0^{\pi / 4} \frac{\sin x+\cos x}{9+16 \sin 2 x} \)dx
Solution:
Let I = \(\int_0^{\pi / 4} \frac{\sin x+\cos x}{9+16 \sin 2 x} \)dx
Put sin x – cos x = t
⇒ (cos x + sin x) dx = dt (sin x – cos x)² = t²
⇒ sin² x + cos² x – 2 sin x cos x = t²
⇒ 1 – sin2x = t² 1 – t² = sin 2x
LL: x = 0 then t = sin 0 – cos 0 = -1

Question 7.
\(\int_0^{\frac{\pi}{4}} \frac{\sin x \cos x}{\cos ^4 x+\sin ^4 x} d x\)
Solution:
Let I = \(\int_0^{\frac{\pi}{4}} \frac{\sin x \cos x}{\cos ^4 x+\sin ^4 x} d x\)
= \(=\int_0^{\frac{\pi}{4}} \frac{\frac{(\sin x \cos x)}{\cos ^4 x}}{\frac{\left(\cos ^4 x+\sin ^4 x\right)}{\cos ^4 x}} d x\) = \(\int_0^{\frac{\pi}{4}} \frac{\tan x \sec ^2 x}{1+\tan ^4 x} d x\)
Let tan²x = t ⇒ 2 tan x sec² x dx = dt
∴ I = \(\frac{1}{2} \int_0^1 \frac{\mathrm{dt}}{1+\mathrm{t}^2}=\frac{1}{2}\left[\tan ^{-1} \mathrm{t}\right]_0^1\)
= \(\frac{1}{2}\)[tan^-1 1 – tan^-1 0]
= \(\frac{1}{2}\left[\frac{\pi}{4}\right]\)
= \(\frac{\pi}{8}\)

Question 8.
\(\int_0^{\frac{\pi}{2}} \frac{\cos ^2 x}{\cos ^2 x+4 \sin ^2 x} d x\)
Solution:

= [tan^-1(∞) – tan^-1(0)] = \(\frac{\pi}{2}\)
From (1) I = \(-\frac{\pi}{6}+\frac{2}{3}\left[\frac{\pi}{2}\right]=\frac{\pi}{3}-\frac{\pi}{6}=\frac{\pi}{6}\)

Question 9.
\(\int_{\frac{\pi}{2}}^\pi e^x\left(\frac{1-\sin x}{1-\cos x}\right) d x\)
Solution:

Question 10.
\(\int_0^{\frac{\pi}{2}}\) sin2x tan^-1 (sin x) dx
Solution:
Let I = \(\int_0^{\frac{\pi}{2}}\) sin2x tan^-1 (sin x) dx
= \(\int_0^{\frac{\pi}{2}}\) 2 sin x cos x tan^-1 (sin x) dx
[∵ sin 2x = 2 sin x cos x]
Put sin x = t ⇒ cos x dx = dt ⇒ dx = \(\frac{d t}{\cos x}\)
L.L : x = 0 ⇒ t = 0 and

Question 11.
\(\int_1^4\) [|x – 1| + |x – 2| + |x – 3|]dx
Solution:
Let I = \(\int_1^4\) [|x – 1| + |x – 2| + |x – 3|]dx
= \(\int_1^2\) {|x – 1| + |x – 2| + |x – 3|}dx + \(\int_2^3\) {|x – 1| + |x – 2| + |x – 3|}dx + \(\int_3^4\) {|x – 1| + |x – 2| + |x – 3|}dx
= \(\int_1^2\) {(x – 1) + (x – 2) + (x – 3)}dx + \(\int_2^3\) {(x – 1 + x – 2) – (x – 3)}dx + \(\int_3^4\) {x – 1 + x + 2 + x – 3}dx
= \(\int_1^2\)(-x + 4) dx + \(\int_2^3\) x dx + \(\int_3^4\)(3x – 6) dx ……….. (2)
= \(\left[-\frac{x^2}{2}+4 x\right]_1^2+\left[\frac{x^2}{2}\right]_2^3+\left[\frac{3 x^2}{2}-6 x\right]_3^4\)
= [\(\frac{-2^2}{2}\) + 8] – [\(\frac{-1}{2}\) + 4] + \(\frac{1}{2}\)(3² – 2²) + (\(\frac{3}{2}\) × 4² – 6 × 4) – (\(\frac{3}{2}\) × 3² – 6 × 3)
= 6 – \(\frac{7}{2}\) + \(\frac{5}{2}\) + (24 – 24) – (-\(\frac{9}{2}\))
= \(\frac{12-7+5+9}{2}\) = \(\frac{19}{2}\)

Question 12.
\(\int_0^\pi\)log(1 + cos x) dx
Solution:
Let I = \(\int_0^\pi\)log(1 + cos x) dx
[∵ \(\int_0^{\mathrm{a}}\)f(x) dx = \(\int_0^{\mathrm{a}}\)f(a – x) dx]
I = \(\int_0^\pi\)log{1 + cos(π – x)} dx
= \(\int_0^\pi\)log(1 – cos x) dx [∵ cos(π – x) = -cos x]
= \(\int_0^\pi \log \left\{2 \sin ^2\left(\frac{x}{2}\right)\right\}\)dx [∵ 1 – cos x = 2 sin²x/2]
= \(\int_0^\pi\left\{\log 2+2 \log \left(\sin \frac{x}{2}\right)\right\}\)dx [∵ log mn² = log m + 2 log n]
= \(\int_0^\pi \log 2 d x+2 \int_0^\pi \log \left(\sin \frac{x}{2}\right) d x\)
In the second integral, put \(\frac{x}{2}\) = t ⇒ dx = 2 dt, and
limits when x = 0, t = 0 & when x = π, t = π/2
∴ I = \(\log 2(x)_0^\pi+2 \int_0^{\pi / 2} \log (\sin t) 2 d t\)
= (log 2) (π – 0) + 4(-\(\frac{\pi}{2}\) log 2)
[∵ \(\int_0^{\pi / 2}\) log sin x dx = –\(\frac{\pi}{2}\)log 2]]
= π log 2 – 2π log 2 = -π log 2

Question 13.
\(\int_1^2 e^{2 x}\left(\frac{1}{x}-\frac{1}{2 x^2}\right)\) dx
Solution:
Put 2x = t ⇒ 2 dx = dt
When x = 1, t = 2 and when x = 2, t = 4
∴ \(\int_1^2\left(\frac{1}{\mathrm{x}}-\frac{1}{2 \mathrm{x}^2}\right) \mathrm{e}^{2 \mathrm{x}} \mathrm{dx}=\frac{1}{2} \int_2^4\left(\frac{2}{\mathrm{t}}-\frac{2}{\mathrm{t}^2}\right) \mathrm{e}^{\mathrm{t}} \mathrm{dt}\)
= \(\int_2^4\left(\frac{1}{t}-\frac{1}{t^2}\right) e^t d t=\int_2^4 e^t\left(\frac{1}{t}+\left(\frac{1}{t}\right)\right) d t=\left[\frac{e^t}{t}\right]_2^4\)
= \(\frac{e^4}{4}-\frac{e^2}{2}=\frac{e^2\left(e^2-2\right)}{4}\)

Question 14.
If f and g are defined as f(x) = f(a – x) and g(x) + g(a – x) = 4, then show that \(\int_0^{\mathrm{a}}\)f(x) g(x) dx = 2\(\int_0^{\mathrm{a}}\)f(x) dx]
Solution:
Let I = \(\int_0^{\mathrm{a}}\)f(x) g(x) dx [∵ \(\int_0^{\mathrm{a}}\)f(x) dx = \(\int_0^{\mathrm{a}}\)f(a – x) dx]
⇒ I = \(\int_0^{\mathrm{a}}\)f(x) dx = \(\int_0^{\mathrm{a}}\)f(a – x) g(a – x) dx …………. (i)
⇒ I = \(\int_0^{\mathrm{a}}\) f(x) {4 – g(x)} dx ………… (ii)
[∵ f(x) = f(a – x) and g(x) + g(a – x) = 4]
On adding eq (i) & (ii), we get
2I = \(\int_0^{\mathrm{a}}\) 4 f(x) dx
⇒ I = 2\(\int_0^{\mathrm{a}}\) f(x) dx
Hence proved.

Practicing the AP Inter 2nd Year Maths Study Material and AP Inter 2nd Year Maths Chapter 7 Integrals Solutions Exercise 7i will help students to clear their doubts quickly.

Inter 2nd Year Maths Integrals Solutions Exercise 7i

Integrals Exercise 7i Solutions

I. Evaluate the following definite integrals.

Question 1.
\(\int_0^1 \frac{x}{x^2+1}\) dx
Solution:
Let I = \(\int_0^1 \frac{x}{x^2+1}\) dx = \(\frac{1}{2} \int_0^1 \frac{2 x}{x^2+1}\) dx
= \(\frac{1}{2}\left(\log \left|x^2+1\right|\right)_0^1\)
= \(\frac{1}{2}\)(log 2 – log 1) = \(\frac{1}{2}\)log 2

Question 2.
\(\int_0^{\frac{\pi}{2}} \frac{\sin x}{1+\cos ^2 x}\) dx
Solution:
Let I = \(\int_0^{\frac{\pi}{2}} \frac{\sin x}{1+\cos ^2 x}\) dx
Put cos x = t ⇒ -sin x dx = dt
L.L : x = 0 we have t = 1 & U.L : x = \(\frac{\pi}{2}\) then t = 0
\(\int_1^0 \frac{1}{1+t^2}\) (-(dt)) = \(\int_0^1 \frac{1}{1+t^2}\) dt
= \(\left(\tan ^{-1} \mathrm{t}\right)_0^1=\frac{\pi}{4}\)

Question 3.
\(\int_{-1}^1 \frac{d x}{x^2+2 x+5}\)
Solution:
Let I = \(\int_{-1}^1 \frac{d x}{x^2+2 x+5}\) = \(\int_{-1}^1 \frac{d x}{x^2+2 x+1+4}\)
= \(\int_{-1}^1 \frac{\mathrm{dx}}{(\mathrm{x}+1)^2+2^2}\) = \(\frac{1}{2}\left(\tan ^{-1}\left(\frac{x+1}{2}\right)\right)_{-1}^1\)
= \(\frac{1}{2}\left(\tan ^{-1} \frac{2}{2}-\tan ^{-1} \frac{0}{2}\right)\) = \(\frac{\pi}{2} \cdot \frac{\pi}{4}=\frac{\pi}{8}\)

Question 4.
\(\int_0^{\frac{\pi}{4}} 2 \tan ^3 x\) dx
Solution:

Question 5.
\(\int_0^1 x e^x\) dx
Solution:
Let I = \(\int_0^1 x e^x\) dx
= \(\left(x e^x\right)_0^1-\left(e^x\right)_0^1\)
= 1 e – 0 – e¹ + 1 = 1

Question 6.
\(\int_1^2 \log x\) dx
Solution:
Let I = \(\int_1^2 \log x\) dx = \(\int_1^2 \log x .1\) dx
= ∫ logx . dx = ∫[\(\frac{d}{d x}\)(log x) ∫1 dx]dx
= x log x – ∫\(\frac{1}{x}\)x dx = x log x – x
∴ \(\int_1^2 \log x\) dx = \((x \log x)_1^2-(x)_1^2\)
= 2 log 2 – 0 – (2 – 1)
= 2 log 2 – 1

Question 7.
\(\int_0^4 \frac{x^2}{1+x}\) dx
Solution:
Let I = \(\int_0^4 \frac{x^2}{1+x}\) dx = \(\int_0^4 \frac{x^2-1+1}{x+1}\) dx
= \(\int_0^4\left(x-1+\frac{1}{x+1}\right)\) dx = \(\left(\frac{x^2}{2}\right)_0^4-(x)_0^4+[\log (x+1)]_0^4\) dx
= 8 – 0 – 4 + log 5 – 0
= 4 + log 5

II. Evaluate the following definite integrals.

Question 1.
\(\int_0^2 x \sqrt{x+2}\) dx (Put x + 2 = t²)
Solution:

Question 2.
Evaluate \(\int_0^{\frac{\pi}{2}} \sqrt{\sin \phi} \cos ^5 \phi d \phi\)
Solution:
Let I = \(\int_0^{\frac{\pi}{2}} \sqrt{\sin \phi} \cos ^5 \phi d \phi\)
= \(\int_0^{\frac{\pi}{2}} \sqrt{\sin \phi}\left(1-\sin ^2 \phi\right)^2 \cos \phi d \phi\)
Put sin Φ = t ⇒ cosΦ dΦ = dt
L.L Φ = 0, then t = 0, U.L : Φ = \(\frac{\pi}{2}\) then t = 1
∴ I = \(\int_0^1 \sqrt{\mathrm{t}}\left(1-\mathrm{t}^2\right)^2 \mathrm{dt}\) = \(\)
= \(\int_0^1\left(t^{9 / 2}+t^{1 / 2}-2 t^{51 / 2}\right)\) dt = \(\left(\frac{t^{11 / 2}}{\frac{11}{2}}+\frac{t^{3 / 2}}{\frac{3}{2}}-2 \frac{t^{7 / 2}}{\frac{7}{2}}\right)_0^1\)
= \(\frac{2}{11}\) + \(\frac{2}{3}\) – \(\frac{4}{7}\) = \(\frac{42+154-132}{231}\)
= \(\frac{64}{231}\)

Question 3.
\(\int_0^2 \frac{d x}{x+4-x^2}\)
Solution:

Question 4.
\(\int_0^1 \sin ^{-1} x d x\)
Solution:
Let I = \(\int_0^1 \sin ^{-1} x\) dx = \(\left(x \sin ^{-1} x\right)_0^1-\int_0^1 \frac{x}{\sqrt{1-x^2}}\) dx
Put 1 – x² = t ⇒ -2x dx = dt
L.L : x = 0 ⇒ t = 1 & U.L : x = 1 ⇒ t = 0
= (1) sin^-1(1) – 0 + \(\frac{1}{2} \int_0^1 \frac{d t}{\sqrt{t}}\) = \(\frac{\pi}{2}+\frac{1}{2}\left(\frac{t^{1 / 2}}{1 / 2}\right)_0^1\)
= \(\frac{\pi}{2}\) – 1

Question 5.
\(\int_0^1 \sin ^{-1}\left(\frac{2 x}{1+x^2}\right)\) dx
Solution:
Let I = \(\int_0^1 \sin ^{-1}\left(\frac{2 x}{1+x^2}\right)\) dx
Put x = tan θ ⇒ dx = sec² θ dθ ⇒ θ = tan^-1x
L.L : x = 0 then θ = 0 and & U.L: x = 1 then θ = \(\frac{\pi}{4}\)

Question 6.
\(\int_0^1 x \tan ^{-1} x\) dx
Solution:

Practicing the AP Inter 2nd Year Maths Study Material and AP Inter 2nd Year Maths Chapter 7 Integrals Solutions Exercise 7h will help students to clear their doubts quickly.

Inter 2nd Year Maths Integrals Solutions Exercise 7h

Integrals Exercise 7h Solutions

I. Evaluate the following definite integrals

Question 1.
\(\int_{-1}^1(x+1) d x\)
Solution:
Let I = \(\int_{-1}^1(x+1) d x=\left[\frac{x^2}{2}+x\right]_{-1}^1\)
= \(\left[\frac{1}{2}+1\right]\) – \(\left(\frac{(-1)^2}{2}+(-1)\right)\)
= \(\frac{3}{2}+\frac{1}{2}=\frac{4}{2}\) = 2

Question 2.
Evaluate \(\int_2^3 \frac{1}{x} d x\)
Solution:
Let I = \(\int_2^3 \frac{1}{x}\) dx
= \((\log \mathrm{x})_2^3\)
= log 3 – log 2 = log \(\frac{3}{2}\)

Question 3.
\(\int_1^2\left(4 x^3-5 x^2+6 x+9\right) d x\)
Solution:
Let I = \(\int_1^2\left(4 x^3-5 x^2+6 x+9\right) d x\)
\(\left(\frac{4 x^4}{4}-\frac{5 x^3}{3}+\frac{6 x^2}{2}+9 x\right)_1^2=\left(x^4-\frac{5 x^3}{3}+3 x^2+9 x\right)_1^2\)
= \(\left(2^4-\frac{5(2)^3}{3}+3(2)^2+9(2)\right)-\left(1-\frac{5}{3}+3+9\right)\)
= \(16-\frac{40}{3}+12+18-\left(13-\frac{5}{3}\right)=\left(46-\frac{40}{3}\right)-\left(13-\frac{5}{3}\right)\)
= \(\frac{138-40}{3}-\left(\frac{39-5}{3}\right)=\frac{98}{3}-\frac{34}{3}=\frac{64}{3}\).

Question 4.
\(\int_0^\pi 4 \sin 2 x d x\)
Solution:
Let I = \(\int_0^{\frac{\pi}{4}} \sin 2 x d x=\left(\frac{-\cos 2 x}{2}\right)_0^{\pi / 4}\)
= \(\frac{-\cos 2 \frac{\pi}{4}}{2}+\frac{\cos (0)}{2}\)
= 0 +\(\frac{1}{2}\) = \(\frac{1}{2}\)

Question 5.
\(\int_0^\pi 2 \cos 2 x d x\)
Solution:
Let I = \(\int_0^{\frac{\pi}{2}} \cos 2 x d x=\left[\frac{\sin 2 x}{2}\right]_0^{\frac{\pi}{2}}\)
= \(\frac{1}{2}\) (sin π – sin 0) = 0

Question 6.
\(\int_4^5 e^x d x\)
Solution:
Let I = \(\int_4^5 e^x d x=\left[e^x\right]_4^5\)
= e⁵ – e⁴ = e⁴(e – 1)

Question 7.
\(\int_0^\pi 4 \tan x d x\)
Solution:
Let I = \(\int_0^{\frac{\pi}{4}} \tan x d x=[-\log \cos x]_0^{\pi / 4}\)
= – log|cos \( \frac{\pi}{4}\) |+ log |cos 0|
= – log \( \frac{1}{\sqrt{2}}\) + log 1
= – log 2^1/2 + 0 = \( \frac{1}{2}\) log 2

Question 8.
\(\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} {cosec} x d x\)
Solution:
Let I = \(\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} {cosec} x d x=[\log |{cosec} x-\cot x|]_{\pi / 6}^{\pi / 4}\)
= log |cosec \( \frac{\pi}{4}\) – cot \(\frac{\pi}{4}\)| – [log|cosec \(\frac{\pi}{6}\) – cot\(\frac{\pi}{6}\)|]
= \(\log |\sqrt{2}-1|-\log |2-\sqrt{3}|=\log \left(\frac{\sqrt{2}-1}{2-\sqrt{3}}\right)\)

Question 9.
\(\int_0^1 \frac{d x}{\sqrt{1-x^2}}\)
Solution:
Let I = \(\int_0^1 \frac{d x}{\sqrt{1-x^2}}=\left[\sin ^{-1} x\right]_0^1\)
= sin^-1 1 – sin^-1 0
= \(\frac{\pi}{2}\) – 0 = \(\frac{\pi}{2}\)

Question 10.
\(\int_0^1 \frac{d x}{1+x^2}\)
Solution:
Let I = \(\int_0^1 \frac{d x}{1+x^2}=\left[\tan ^{-1} x\right]_0^1\)
= tan^-1 1 – tan^-1 0
= \(\frac{\pi}{4}\) – 0 = \(\frac{\pi}{4}\)

Question 11.
\(\int_2^3 \frac{d x}{x^2-1}\)
Solution:
Let I = \(\int_2^3 \frac{\mathrm{dx}}{\mathrm{x}^2-1}\)
= \(\left[\frac{1}{2} \log \left|\frac{\mathrm{x}-1}{\mathrm{x}+1}\right|\right]_2^3\)
= \(\frac{1}{2}\left[\log \left|\frac{2}{4}\right|-\log \left|\frac{1}{3}\right|\right]\)
= \(\frac{1}{2}\left[\log \frac{1}{2}-\log \frac{1}{3}\right]\) = \(\frac{1}{2} \log \frac{3}{2}\)

Question 12.
\(\int_2^3 \frac{x d x}{x^2+1}\)
Solution:
Let I = \(\int_2^3 \frac{x}{x^2+1}\) dx
= \(\frac{1}{2} \int_2^3 \frac{2 x}{x^2+1}\) dx = \(\frac{1}{2}\left[\log \left(1+x^2\right)\right]_2^3\)
= \(\frac{1}{2}\)[log 10 – log 5] = \(\frac{1}{2}\) log \(\frac{10}{5}\)
= \(\frac{1}{2}\) log 2

Question 13.
\(\int_0^1 x e^{x^2} d x\)
Solution:
Let I = \(\int_0^1 x e^{x^2}\) dx
Put x² = t ⇒ 2x dx = dt ⇒ x dx = \(\frac{d t}{2}\)
L.L. x = 0 then t = 0 & U.L. x = 1 then t = 1
∴ I = \(\frac{1}{2} \int_0^1 e^t d t=\frac{1}{2}\left[e^t\right]_0^1\)
= \(\frac{1}{2}\) (e – 1)

Question 14.
\(\int_0^{\pi / 4}\left(2 \sec ^2 x+x^3+2\right)\) dx
Solution:
Let I = \(\int_0^{\pi / 4}\left(2 \sec ^2 x+x^3+2\right)\) dx
= \(2 \int_0^{\frac{\pi}{4}} \sec ^2 x d x+\int_0^{\frac{\pi}{4}} x^3 d x+2 \int_0^{\frac{\pi}{4}} 1 d x\)
= \(2(\tan x)_0^{\pi / 4}+\frac{1}{4}\left(x^4\right)_0^{\pi / 4}+2(x)_0^{\pi / 4}\)
= \(2 \tan \frac{\pi}{4}+\frac{1}{4}\left(\frac{\pi}{4}\right)^4+2\left(\frac{\pi}{4}\right)=2+\frac{\pi}{2}+\frac{\pi^4}{1024}\)

Question 15.
\(\int_0^\pi\left(\sin ^2 \frac{x}{2}-\cos ^2 \frac{x}{2}\right) d x\)
Solution:
Let I = \(\int_0^\pi\left(\sin ^2 \frac{x}{2}-\cos ^2 \frac{x}{2}\right)\) dx
= –\(\int_0^\pi\left(\cos ^2 \frac{x}{2}-\sin ^2 \frac{x}{2}\right)\) dx
= \(-\int_0^\pi \cos x\) d x
= \(-[\sin x]_0^\pi\)
= -(sin π – sin 0) = 0

Question 16.
\(\int_0^{\frac{\pi}{2}} \cos ^2 x \)dx
Solution:
Let I = \(\int_0^{\pi / 2}\left(\frac{\cos 2 x+1}{2}\right)\) dx
= \(\frac{1}{2}\left[\left(\frac{\sin 2 x}{2}\right)_0^{\pi / 2}+(x)_0^{\pi / 2}\right]\)
= \(\frac{1}{2}\)[sin π – sin 0 + \(\frac{\pi}{2}\) – 0]
= \(\frac{1}{2}\)(0 – 0 + \(\)) = \(\frac{\pi}{4}\)

Question 17.
\(\int \frac{1}{\sqrt{1+x}-\sqrt{x}} d x\)
Solution:
Let I = \(\int \frac{1}{\sqrt{1+x}-\sqrt{x}} d x\)
= \(\int_0^1(\sqrt{1+x}+\sqrt{x}) d x=\left(\frac{(1+x)^{3 / 2}}{3 / 2}\right)_0^1+\left(\frac{x^{3 / 2}}{3 / 2}\right)_0^1\)
= \(\frac{2}{3}\) [2^3/2 – 1 + (1 – 0)]
= \(\frac{2}{3}\) (2^3/2) = \(\frac{2}{3}(2 \sqrt{2})\)
= \(\frac{4 \sqrt{2}}{3}\)

Question 18.
\(\int_0^{\frac{\pi}{2}} \sin ^3 x d x\)
Solution:
Let I = \(\int_0^{\frac{\pi}{2}} \sin ^3 x d x\)
I = \(\int_0^{\pi / 2}\left(\frac{3 \sin x-\sin 3 x}{4}\right) d x=\frac{1}{4}\left[\int_0^{\pi / 2} 3 \sin x d x-\int_0^{\pi / 2} \sin 3 x d x\right]\)
= \(\frac{1}{4}\left(-3 \cos x+\frac{\cos 3 x}{3}\right)_0^{\pi / 2}\)
= \(\frac{1}{4}\left(3-\frac{1}{3}\right)\) = \(\frac{2}{3}\)

II. Evaluate the following definite integrals

Question 1.
\(\int_0^1 \frac{2 x+3}{5 x^2+1}\) dx
Solution:

Question 2.
Evaluate \(\int_0^2 \frac{6 x+3}{x^2+4}\) dx
Solution:

Question 3.
Evaluate \(\int_1^2 \frac{5 x^2}{x^2+4 x+3}\) dx
Solution:
Let I = \(\int_1^2 \frac{5 x^2}{x^2+4 x+3}\) dx
I = \(\int_1^2\left[5-\frac{20 x+15}{x^2+4 x+3}\right]\) dx
Let \(\left(\frac{20 x+15}{x^2+4 x+3}\right)=\frac{A}{x+1}+\frac{B}{x+3}\)
20x + 15 = A(x + 3) + B(x + 1)
20 = A + B & 15 = 3A + B ⇒ A = -5/2 & B = 45/2
= \(5 \int_1^2 \mathrm{dx}+\frac{5}{2} \int_1^2 \frac{1}{\mathrm{x}+1} \mathrm{dx}-\frac{45}{2} \int_1^2 \frac{1}{\mathrm{x}+3} \mathrm{dx}\)
= \(\left(5 x+\frac{5}{2} \log (x+1)-\frac{45}{2} \log |x+3|\right)_1^2\)
= 10 + \(\frac{5}{2}\) log|3| – \(\frac{45}{2}\) log 5 – 5 – \(\frac{5}{2}\) log |2| + \(\frac{45}{2}\) log|4|
= 5 + \(\frac{5}{2}\) log|\(\frac{3}{2}\)| – \(\frac{45}{2}\) log|\(\frac{5}{4}\)|
= 5 – \(\frac{5}{2}\)(9 log \(\frac{5}{4}\) – log\(\frac{3}{2}\))

Question 4.
\(\int_0^1\left[x e^x+\sin \frac{\pi x}{4}\right]\) dx
Solution:
Let I = \(\int_0^1\left[x e^x+\sin \frac{\pi x}{4}\right]\) dx
= \(\int_0^1 x e^x d x+\int_0^1 \sin \frac{\pi x}{4} d x\) = \(\left(x e^x-e^x\right)_0^1-\frac{4}{\pi}\left(\cos \frac{\pi x}{4}\right)_0^1\)
= (e – e) – (0 – e⁰) – \(\frac{4}{\pi}\left(\cos \frac{\pi}{4}-\cos 0\right)\)
= 0 + 1 – \(\frac{4}{\pi}\left(\frac{1}{\sqrt{2}}-1\right)\)
= 1 – \(\frac{4}{\pi}\left(\frac{1-\sqrt{2}}{\sqrt{2}}\right) \times \frac{\sqrt{2}}{\sqrt{2}}\) = 1 + \(\frac{2}{\pi}(2-\sqrt{2})\)
= 1 + \(\frac{4}{\pi}-\frac{2 \sqrt{2}}{\pi}\)

Question 5.
\(\int_1^3 \frac{d x}{x^2(x+1)}\)
Solution:
Let, \(\frac{1}{x^2(x+1)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x+1}\) ……….. (1)
1 = Ax(x + 1) + B(x + 1) + C(x²) ………… (2)
Put x = 0, -1 on eqn (2) then
1 = B(0 + 1) ⇒ B = 1 & 1 = (-1)²C ⇒ C = 1
⇒ 1 = Ax² + Ax + Bx + B + Cx²
Comparing the coefficients of x² on both sides,
A + C = 0 ⇒ A = -C = -1
Substituting A, B, C values in (1) then
\(\frac{1}{x^2(x+1)}=\frac{-1}{x}+\frac{1}{x^2}+\frac{1}{(x+1)}\)
\(\int_1^3 \frac{1}{x^2(x+1)} d x=\int_1^3\left(\frac{-1}{x}+\frac{1}{x^2}+\frac{1}{x+1}\right) d x\)
= \(\left(-\log |x|-\frac{1}{x}+\log |x+1|\right)_1^3=\left(\log \left|\frac{x+1}{x}\right|-\frac{1}{x}\right)_1^3\)
= \(\left(\log \left|\frac{4}{3}\right|-\frac{1}{3}\right)-\left(\log \left|\frac{2}{1}\right|-1\right)\)
= log \(\frac{4}{3}\) – log 2 + 1 – \(\frac{1}{3}\)
= log \(\frac{2}{3}\) + \(\frac{2}{3}\) = \(\frac{2}{3}\) + log\(\frac{2}{3}\)

Question 6.
\(\int_0^1 \frac{x^{\frac{1}{4}}}{1+x^{\frac{1}{2}}}\) dx
Solution:
Let I = \(\int_0^1 \frac{x^{\frac{1}{4}}}{1+x^{\frac{1}{2}}}\) dx
Put x = t² ⇒ dx = 2t dt ⇒ x^1/2 = t; x^1/4 = x^1/2

Practicing the AP Inter 2nd Year Maths Study Material and AP Inter 2nd Year Maths Chapter 7 Integrals Solutions Exercise 7g will help students to clear their doubts quickly.

Inter 2nd Year Maths Integrals Solutions Exercise 7g

Integrals Exercise 7g Solutions

I. Integrate the following functions.

Question 1.
\(\sqrt{4-x^2}\)
Solution:
Let I = \(\int \sqrt{4-x^2}\) dx
= \(\int \sqrt{(2)^2-(x)^2}\) dx [∵ \(\int \sqrt{a^2-x^2} d x=\frac{x}{2} \sqrt{a^2-x^2}+\frac{a^2}{2} \sin ^{-1} \frac{x}{a}+C\)]
∴ I = \(\frac{x}{2} \sqrt{4-x^2}+2 \sin ^{-1} \frac{x}{2}+C\)

Question 2.
\(\sqrt{1-4 x^2}\)
Solution:
Let I = \(\int \sqrt{1-4 \mathrm{x}^2} \mathrm{dx}\)
= ∫ \(\sqrt{4\left(\frac{1}{4}-x^2\right)}\) dx = 2 ∫\(\sqrt{\left(\left(\frac{1}{2}\right)^2-x^2\right)}\) dx
= \(2 \cdot \frac{x}{2} \sqrt{\frac{1}{4}-x^2}+\frac{1}{4} \cdot \frac{2}{2} \sin ^{-1} \frac{x}{1 / 2}+C\)
[∵ \(\int \sqrt{a^2-x^2} d x=\frac{x}{2} \sqrt{a^2-x^2}+\frac{a^2}{2} \sin ^{-1} \frac{x}{2}+C\)]
∴ I = \(\frac{x}{2} \sqrt{1-4 x^2}+\frac{1}{4} \sin ^{-1} 2 x+C\)

Question 3.
\(\sqrt{x^2+4 x+6}\)
Solution:

Question 4.
\(\sqrt{x^2+4 x+1}\)
Solution:
Let I = ∫ \(\sqrt{x^2+4 x+1} d x=\int \sqrt{\left(x^2+4 x+4\right)-3}\) dx
= ∫\(\sqrt{x^2+4 x+1-2^2+2^2}\) dx = ∫\(\sqrt{(x+2)^2-(\sqrt{3})^2}\) dx
∴ I = \(\left(\frac{x+2}{2}\right) \sqrt{x^2+4 x+1}-\frac{3}{2} \log (x+2)+\sqrt{x^2+4 x+1}\) + C
[∵ \(\int \sqrt{x^2-a^2} d x=\frac{x}{2} \sqrt{x^2-a^2}-\frac{a^2}{2} \log \left|x+\sqrt{x^2-a^2}\right|\)]

Question 5.
\(\sqrt{1-4 x-x^2}\)
Solution:
Let I = ∫\(\sqrt{1-4 x-x^2}\) dx = ∫\(\sqrt{-\left(x^2+4 x-1-2^2+2^2\right)}\) dx
= ∫\(-\left[(x+2)^2-(\sqrt{5})^2\right]\) dx
= ∫\(\sqrt{(\sqrt{5})^2-(x+2)^2}\) dx
∴ I = \(\frac{(x+2)}{2} \sqrt{1-4 x-x^2}+\frac{5}{2} \sin ^{-1}\left(\frac{x+2}{\sqrt{5}}\right)+C\)
[∵ \(\int \sqrt{a^2-x^2} d x=\frac{x}{2} \sqrt{a^2-x^2}+\frac{a^2}{2} \sin ^{-1} \frac{x}{a}+C\)]

Question 6.
\(\sqrt{x^2+4 x-5}\)
Solution:
Let I = ∫\(\sqrt{x^2+4 x-5-2^2}+2^2\) dx
= ∫\(\sqrt{(x+2)^2-5-4}\) dx = ∫\(\sqrt{(x+2)^2-(3)^2}\) dx
∴ I = \(\frac{(x+2)}{2} \sqrt{x^2+4 x-5}-\frac{9}{2} \log \left|(x+2)+\sqrt{x^2+4 x-5}\right|\) + C
[∵ \(\int \sqrt{x^2-a^2} d x=\frac{x}{2} \sqrt{x^2-a^2}-\frac{a^2}{2} \log \left|x+\sqrt{x^2-a^2}\right|\)]

Question 7.
\(\sqrt{1+3 x-x^2}\)
Solution:

Question 8.
\(\sqrt{x^2+3 x}\)
Solution:

Question 9.
\(\sqrt{1+\frac{x^2}{9}}\)
Solution:

Practicing the AP Inter 2nd Year Maths Study Material and AP Inter 2nd Year Maths Chapter 7 Integrals Solutions Exercise 7f will help students to clear their doubts quickly.

Inter 2nd Year Maths Integrals Solutions Exercise 7f

Integrals Exercise 7f Solutions

I. Integrate the functions.

Question 1.
x sin x
Solution:
∫x sin x dx
By parts
∫ f(x) g(x) dx = f(x) ∫ g(x) dx – ∫ [f'(x) ∫ g(x) dx] dx
According to ILATE taking x as first function & second function sin x apply by parts
= x ∫ sin x dx – ∫ [1 ∫ (sin x dx)]dx
= -x cos x + ∫ cosx dx = -x cos x + sin x + C

Question 2.
x sin 3x
Solution:
Let I = ∫x sin 3x dx
According to ILATE taking x as first function & sin 3x as second function integrating by parts, we get
I = \(x \int \sin 3 x d x-\int\left[\left(\frac{d}{d x} x\right) \int \sin 3 x d x\right] d x\)
= \(\frac{-x \cos 3 x}{3}+\int \frac{\cos 3 x}{3} d x\) [∵ ∫ sin ax dx = \(\frac{-\cos a x}{a}\)]
⇒ I = \(\frac{-x \cos 3 x}{3}+\frac{1}{9} \sin 3 x\) + C

Question 3.
x log x
Solution:
Let I = ∫x log x dx
According to ILATE taking log x as first function and x as second function & integrating by parts, we get,
I = \(\log x \int x d x-\int\left[\left(\frac{d}{d x} \log x\right) \int x d x\right] d x\)
= \(\frac{x^2 \log x}{2}-\frac{1}{2} \int \frac{1}{x} x^2 d x=\frac{x^2 \log x}{2}-\frac{1}{2} \int x d x\)
= \(\frac{x^2 \log x}{2}-\frac{x^2}{4}\) + C

Question 4.
x log 2x
Solution:
Let I = ∫x log 2x dx
According to ILATE, taking log 2x as first function & x as second function and integrating by parts, we get,
I = \(\log 2 x \int x d x-\int\left[\left(\frac{d}{d x} \log 2 x\right) \int x d x\right] d x\)
= \(\frac{x^2}{2} \log 2 x-\int\left(\frac{1}{2 x} \times 2 \times \frac{x^2}{2}\right) d x\) = \(\frac{x^2 \log 2 x}{2}-\frac{1}{2} \int x d x\)
= \(\frac{x^2}{2} \log 2 x-\frac{x^2}{4}\) + C

Question 5.
x² log x
Solution:
Let I = ∫x² log x dx
According to ILATE. taking log x as first function and x² as second function and integrating by parts, we get
I = \(\log x \int x^2 d x-\int\left[\left(\frac{d}{d x} \log x\right) \int x^2 d x\right] d x\)
= \(\frac{x^3}{3} \log x-\int\left(\frac{1}{x} \cdot \frac{x^3}{3}\right) d x=\frac{x^3}{3} \log x-\frac{1}{3} \int x^2 d x\)
= \(\frac{x^3}{3} \log x-\frac{x^3}{9}\) + C

Question 6.
x sec² x
Solution:
Let I = ∫x sec² x dx
According to ILATE. taking x as first function and sec² x as second function and integrating by parts, we get,
I = \(x \int \sec ^2 x d x-\int\left[\left(\frac{d}{d x} x\right) \int \sec ^2 x d x\right] d x\)
= x tan x – ∫1. tan xdx = x tanx – log|sec x| + C
log|sec x| = log \(\left|\frac{1}{\cos x}\right|\) = log 1 – log|cos x|
= -log |cos x| (∵ log 1 = 0)
⇒ I = x tan x + log |cos x| + C

Question 7.
tan^-1 x
Solution:
Let I = ∫1.tan^-1 x dx
⇒ I = \(\tan ^{-1} x \int 1 d x-\int\left[\left(\frac{d}{d x} \tan ^{-1} x\right) \int 1 . d x\right] d x\)
= x tan^-1 x – ∫ \(\frac{1}{1+x^2}\) dx
Let 1 + x² = t ⇒ 2x = \(\frac{d t}{d x}\) ⇒ \(\frac{d t}{2 x}\) = dx
∴ I = x tan^-1 x – ∫ \(\frac{x}{t} \cdot \frac{d t}{2 x}\) = x tan^-1 x – \(\frac{1}{2}\) ∫ \(\frac{1}{t}\) dt
= x tan^-1x – \(\frac{1}{2}\) log |t| + C
= x tan^-1x – \(\frac{1}{2}\) log |1 + x²| + C

Question 8.
(x² + 1)log x
Solution:
Let I = ∫(x² + 1) log x dx
According to ILATE, taking log x as first function & (x² + 1) as second function and integrating by parts we get
I = log x ∫ (x² + 1) dx – \(\int\left[\frac{d}{d x}(\log x) \int\left(x^2+1\right) d x\right] d x\)
⇒ I = log x (\(\frac{x^3}{3}\) + x) – ∫ \(\frac{1}{x}\)(\(\frac{x^3}{3}\) + x) dx
= (\(\frac{x^3}{3}\) + x) log x – ∫ (\(\frac{x^3}{3}\) + 1) dx
= (\(\frac{x^3}{3}\) + x) log x – \(\frac{x^3}{9}\) – x + C

Question 9.
e^x(sinx + cosx)
Solution:
Let I = ∫ e^x(sinx + cosx) dx
Let f(x) = sin x ⇒ f'(x) = cos x then,
I = ∫ e^x [f(x) + f'(x)] dx
We know that ∫ e^x [f(x) + f'(x)] dx = e^x dx
∴ I = e^xsin x + C

Question 10.
\(\left(\frac{1}{x}-\frac{1}{x^2}\right)\)
Solution:
Let I = ∫e^x\(\left(\frac{1}{x}-\frac{1}{x^2}\right)\)dx
Put f(x) = \(\frac{1}{x}\) ⇒ f'(x) = –\(\frac{1}{x^2}\)
∫ e^x [f(x) + f'(x)] dx = e^x f(x)
∴ I = \(\frac{e^x}{x}\) + C

II. Integrate the following functions.

Question 1.
x²e^x
Solution:
Let I = ∫ x²e^x dx
According to ILATE, taking x² as first function & e^x as second function and integrating by parts, we get
I = \(x^2 \int e^x d x-\int\left[\left(\frac{d}{d x} x^2\right) \int e^x d x\right] d x=x^2 e^x-\int 2 x e^x d x=x^2 e^x-2 \int x e^x d x\)
= x²e^x – ∫ [2xe^x dx]
Again, integrating by parts, we get
I = \(x^2 e^x-\left\{2 x \int e^x d x-2 \int\left(\frac{d}{d x}(x) \int e^x d x\right) d x\right\}\)
= x²e^x – 2xe^x + 2 ∫ e^x dx
= x²e^x – 2xe^x + 2e^x + C
∴ I = e^x(x² – 2x + 2) + C

Question 2.
x sin^-1 x
Solution:
Let I = ∫ xsin^-1x dx
According to ILATE, taking sin^-1 x as first function & x as second function and integrating by parts, we get,

Question 3.
x tan^-1 x
Solution:
Let I = ∫ x tan^-1 xdx
According to ILATE, taking tan^-1 x as first [unction & x as second function and integrating by parts, we get,

Question 4.
x cos^-1 x
Solution:
Let I = ∫ x cos^-1 xdx
Put cos^-1 x = t ⇒ x = cos t ⇒ dx = – sin t dt
I = ∫ x cos^-1 x dx = – ∫ t cos t . sin t dt
= –\(\frac{1}{2}\) ∫ t . 2 sin t cos t dt = – \(\frac{1}{2}\) ∫ t sin 2t dt
[∵ 2 sin x cos x = sin 2x]
= \(\frac{1}{4}\) t cos 2t – \(\frac{1}{4} \frac{\sin 2 t}{2}\) + C
= \(\frac{1}{4}\) t cos 2t – \(\frac{1}{8}\)sin 2t + C
= \(\frac{1}{4}\) t (2cot² t – 1) – \(\frac{1}{8}\)(2 sin t cos t) + C
[∵ cos 2x = 2 cos² x – 1 & sin 2x = 2 sin x cos x]
= \(\frac{1}{4}\) t (2cos²t – 1) – \(\frac{1}{4}\) (1 – cos² t)^1/2cos t + C
[∵ sin x = \(\sqrt{1-\cos ^2 x}\)]
∴ ∫ x cos^-1x dx = \(\frac{1}{4}\) cos^-1x(2x² – 1) – \(\frac{1}{4}\)(1 – x²)^1/2 x + C
[put cos^-1 x = t and cos t = x]
= \(\frac{1}{4}\)(2x² – 1)cos^-1x – \(\frac{x}{4} \sqrt{1-x^2}\) + C

Question 5.
\(\frac{x \cos ^{-1} x}{\sqrt{1-x^2}}\)
Solution:
Let I = \(\int \frac{x \cos ^{-1} x}{\sqrt{1-x^2}} d x \Rightarrow I=\int \cos ^{-1} x \frac{x}{\sqrt{1-x^2}} d x\)
According to ILATE consider cos^-1 x as first function & \(\frac{x}{\sqrt{1-x^2}}\) as second function & integrating by parts, we get

Question 6.
x(log x)²
Solution:
Let I = ∫ x(log x)² dx
According to ILATE taking (log x)² as first function & x as second function & integrating by parts, we get

Question 7.
\(\frac{x e^x}{(1+x)^2}\)
Solution:
Let I = ∫\(\frac{x e^x}{(1+x)^2}\) dx = ∫\(\mathrm{e}^{\mathrm{x}} \frac{(\mathrm{x}+1-1)}{(1+\mathrm{x})^2}\) dx
∫e^x\(\left[\frac{1}{(1+x)}-\frac{1}{(1+x)^2}\right]\) dx
Let f(x) = \(\frac{1}{(1+x)}\) ⇒ f'(x) = –\(\frac{1}{1+x)^2}\)
We know that ∫e^x [f(x) + f'(x)] dx = e^xf(x)
⇒ I = ∫e^x\(\left\{\frac{1}{1+x}-\frac{1}{(1+x)^2}\right\}\) dx
= \(\frac{e^x}{1+x}\) + C

Question 8.
\(e^x\left(\frac{1+\sin x}{1+\cos x}\right)\)
Solution:
∫e^x\(\left(\frac{1}{1+\cos x}+\frac{\sin x}{1+\cos x}\right)\) dx
= ∫e^x\(\left(\frac{1}{2 \cos ^2 \frac{x}{2}}+\frac{2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos ^2 \frac{x}{2}}\right)\) dx
[∵ cos 2x = 2 cos²x – 1 & sin 2x = 2 sin x cos x]
= ∫e^x\(\left(\frac{\sec ^2 \frac{x}{2}}{2}+\tan \frac{x}{2}\right)\)dx = ∫e^x\(\left.\tan \frac{x}{2}+\frac{1}{2} \sec ^2 \frac{x}{2}\right)\)dx
Let f(x) = tan\(\frac{x}{2}\) ⇒ f'(x) = \(\frac{\sec ^2 \frac{x}{2}}{2}\)
∴ ∫e^x\(\left(\frac{1+\sin x}{1+\cos x}\right)\) dx = e^x tan\(\frac{x}{2}\) + C
[∵ ∫ e^x[f(x) + f'(x)] dx = e^xf(x)]

Question 9.
\(\frac{(x-3) e^x}{(x-1)^3}\)
Solution:
Let I = \(\int e^x\left[\frac{x-3}{(x-1)^3}\right] d x=\int e^x\left[\frac{x-1-2}{(x-1)^3}\right]\) dx
= \(\int \mathrm{e}^{\mathrm{x}}\left(\frac{\mathrm{x}-1}{(\mathrm{x}-1)^3}-\frac{2}{(\mathrm{x}-1)^3}\right) \mathrm{dx}=\int \mathrm{e}^{\mathrm{x}}\left(\frac{1}{(\mathrm{x}-1)^2}-\frac{2}{(\mathrm{x}-1)^3}\right) \mathrm{dx}\)
Let f(x) = \(\frac{1}{(x-1)^2}\) ⇒ f'(x) = \(\frac{-2}{(x-1)^3}\)
∴ I = \(\frac{\mathrm{e}^x}{(\mathrm{x}-1)^2}\) + C
[∵ ∫ e^x[f(x) + f'(x)] dx = e^xf(x)]

Question 10.
e^2x sin x
Solution:
Let I = e^2x sin x dx
According to ILATE, taking sin x as first function & e^2x as second function & integrating by parts, we get

According to ILATE, taking sin x as first function & e^2x as second function & integrating by parts, we get

Put value of I₁ in eq(1), we get

Question 11.
\(\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)\)
Solution:

Question 12.
\(\frac{2+\sin 2 x}{1+\cos 2 x} e^x\)
Solution:
Let I = ∫\(\frac{2+\sin 2 x}{1+\cos 2 x} e^x\) dx
= ∫\(\left(\frac{2}{1+\cos 2 x}+\frac{\sin 2 x}{1+\cos 2 x}\right)\) e^x dx
= ∫\(\left(\frac{2}{2 \cos ^2 x}+\frac{2 \sin x \cos x}{2 \cos ^2 x}\right)\) e^x dx
= ∫\(\left(\frac{1}{\cos ^2 x}+\tan x\right)\) e^x dx
= ∫(sec² + tan x)e^x dx = e^xtan x + C
[∵ ∫ e^x[f(x) + f'(x)] dx = e^xf(x)]

III. Integrate the functions

Question 1.
\(\tan ^{-1} \sqrt{\frac{1-x}{1+x}}\)
Solution:

Question 2.
\(\frac{\sqrt{x^2+1}\left[\log \left(x^2+1\right)-2 \log x\right]}{x^4}\)
Solution:

Question 3.
(sin^-1 x)²
Solution:
I = ∫(sin^-1 x)² dx
Put sin^-1 x = θ ⇒ x = sin θ ⇒ dx = cos θ dθ
= ∫θ²cos θ dθ
= ∫θ² cos θ dθ – ∫2θ (∫cosθ dθ) dθ
= θ² – 2 ∫θ sinθ dθ
= θ² sin θ – 2(θ ∫ sinθ dθ – ∫1(∫sinθ dθ)) dθ
= θ²sin θ – 2(-θ cosθ + ∫cosθ dθ)
= θ²sinθ + 2θ cosθ – 2 sinθ + C
= (sin^-1x)²x + 2sin^-1x\(\sqrt{1-\sin ^2 \theta}\) – 2x + C
= x(sin^-1x)² + 2\(\sqrt{1-x^2}\) sin^-1x – 2x + C

Practicing the AP Inter 2nd Year Maths Study Material and AP Inter 2nd Year Maths Chapter 7 Integrals Solutions Exercise 7e will help students to clear their doubts quickly.

Inter 2nd Year Maths Integrals Solutions Exercise 7e

Integrals Exercise 7e Solutions

I. Integrate the Rational Functions.

Question 1.
\(\frac{x}{(x+1)(x+2)}\)
Solution:
∫\(\frac{x}{(x+1)(x+2)}\) dx
Let \(\frac{\mathrm{x}}{(\mathrm{x}+1)(\mathrm{x}+2)}\)
= \(\frac{\mathrm{A}}{(\mathrm{x}+1)}+\frac{\mathrm{B}}{(\mathrm{x}+2)}\)
⇒ \(\frac{x}{(x+1)(x+2)}\) = \(\frac{\mathrm{A}(\mathrm{x}+2)+\mathrm{B}(\mathrm{x}+1)}{(\mathrm{x}+1)(\mathrm{x}+2)}\)
⇒ x = x(A + B) + 2A + B
On equating the coefficient of x and constant terms on both sides, we get
A + B = 1 ………. (1) & 2A + B = 0 ………….. (2)
(2) – (1) ⇒ A = -1
Put the value of A in eq(1), we get
-1 + B = 1 ⇒ B = 2
∴ ∫\(\frac{x}{(x+1)(x+2)}\) dx = ∫\(\frac{-1}{(x+1)}\) dx + ∫\(\frac{2}{(x+2)}\) dx
= log(x + 2)² – log(x + 1) + C [∵ ∫\(\frac{1}{x}\) dx = log|x| + C]
= \(\log \frac{(x+2)^2}{|x+1|}+C\) [∵ loga – logb = log \(\frac{\mathrm{b}}{\mathrm{a}}\)]

Question 2.
\(\frac{1}{x^2-9}\)
Solution:
∫\(\frac{1}{x^2-9}\) dx
= ∫\(\frac{1}{x^2-3^2}\) dx = ∫\(\frac{1}{(x+3)(x-3)}\) dx
Let \(\frac{1}{(x+3)(x-3)}=\frac{A}{(x+3)}+\frac{B}{(x-3)}\)
⇒ 1 = A(x – 3) + B(x + 3) ⇒ 1 = x(A + B) + (-3A + 3B)
On equating the coefficient of x and constant terms on both sides, we get
A + B = 0 and -3A + 3B = 1
On solving, we get A = –\(\frac{1}{6}\) and B = \(\frac{1}{6}\)
∴ ∫\(\frac{1}{(x+3)(x-3)}\) dx = \(\frac{-1}{6(x+3)}\) dx + \(\frac{1}{6(x-3)}\) dx
= \(-\frac{1}{6} \log |x+3|+\frac{1}{6} \log |x-3|+C\)
= \(\frac{1}{6} \log \left|\frac{x-3}{x+3}\right|+C\) [∵ loga – logb = log \(\frac{\mathrm{b}}{\mathrm{a}}\)]

Question 3.
\(\frac{1-x^2}{x(1-2 x)}\)
Solution:
Let ∫\(\frac{1-x^2}{x(1-2 x)}\) dx
Here, degree of numerator is equal to degree of denominator, so divide the numerator by denominator.

On comparing the coefficient of x and constant terms on both sides, we get
2A + B = \(\frac{1}{2}\) and -A = -1 ⇒ A = 1
⇒ 2 × 1 + B = \(\frac{1}{2}\) ⇒ B = \(\frac{1}{2}\) – 2 = \(\frac{-3}{2}\)
∴ I₂ = ∫\(\left[\frac{1}{x}-\frac{3}{2(2 x-1)}\right]\) dx = ∫\(\frac{1}{x} d x-\frac{3}{2} \int \frac{1}{2 x-1}\) dx
⇒ I₂ = log x – \(\frac{3}{2} \frac{\log |2 x-1|}{2}\) + C₂
[∵ ∫\(\frac{1}{x}\) dx = log x + C]
Thus, on putting the values of I₁ and I₂ in eq(1), we get
I = \(\frac{1}{2}\)x + log x – \(\frac{3}{4}\) log|2x – 1| + C [∵ C₁ + C₂ = C]

II. Integrate the Rational Functions.

Question 1.
\(\frac{3 x-1}{(x-1)(x-2)(x-3)}\)
Solution:
Let \(\frac{3 x-1}{(x-1)(x-2)(x-3)}=\frac{A}{(x-1)}+\frac{B}{(x-2)}+\frac{C}{(x-3)}\)
⇒ \(\frac{3 x-1}{(x-1)(x-2)(x-3)}\)
⇒ \(\frac{A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2)}{(x-1)(x-2)(x-3)}\)
⇒ 3x – 1 = A[x² – 5x + 6] + B[x² – 4x + 3] + C[x² – 3x + 2]
⇒ 3x – 1 = x²(A + B + C) + x(-5A – 4B – 3C) + (6A + 3B + 2C)
On equating the coefficients of x², x and constant terms on both sides, we get
A + B + C = O
-5A – 4B – 3C = 3 ………… (ii)
6A + 3B + 2C = -1 ……….. (iii)
From eq(i), we get A = – (B + C)
On putting the value of A in eq (ii) and (iii), we get
-5{-(B + C)} – 4B – 3C = 3
⇒ 5B + 5C – 4B – 3C = 3 = B + 2C = 3 ……….. (iv)
and 6{-(B + C)} + 3B + 2C = -1
⇒ -6B – 6C + 3B + 2C = -1 – 3B – 4C = -1 ………. (v)
On solving eqs (iv) and (v), we get C = 4
On putting the value of C in eq(iv), we get
B + 2 × 4 = 3 ⇒ B = -5
Putting the value of B and C in eq(i), we get
A +(-5) + 4 = 0 ⇒ A = 1
∴ A = 1, B = -5, C = 4
Now, ∫\(\frac{3 x-1}{(x-1)(x-2)(x-3)}\) dx [∵ ∫\(\frac{1}{x}\) dx = log|x| + C]
= ∫\(\left(\frac{\mathrm{A}}{(\mathrm{x}-1)}+\frac{\mathrm{B}}{(\mathrm{x}-2)}+\frac{\mathrm{C}}{(\mathrm{x}-3)}\right)\) dx
= ∫\(\frac{1}{(x-1)}\) dx + ∫\(\frac{(-5)}{(x-2)}\) dx + ∫\(\frac{4}{x-3)}\) dx
= log|x – 1| – 5 log|x – 2| + 4 log|x – 3| + C

Question 2.
\(\frac{x}{(x-1)(x-2)(x-3)}\)
Solution:
∫\(\frac{x}{(x-1)(x-2)(x-3)}\) dx
Let \(\frac{\mathrm{x}}{(\mathrm{x}-1)(\mathrm{x}-2)(\mathrm{x}-3)}\) = \(\frac{\mathrm{A}}{(\mathrm{x}-1)}+\frac{\mathrm{B}}{(\mathrm{x}-2)}+\frac{\mathrm{C}}{(\mathrm{x}-3)}\)
⇒ \(\frac{x}{(x-1)(x-2)(x-3)}\) = \(\frac{A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2)}{(x-1)(x-2)(x-3)}\)
⇒ x = A[x² – 5x + 6] + B[x² – 4x + 3] + C[x² – 3x + 2]
⇒ x = x² (A + B + C) + x(-5A – 4B – 3C)+ (6A + 3B + 2C)
On comparing the coefficients of x², x and constant terms on both sides, we get
A + B + C = 0 ………… (1)
-5A – 4B – 3C = 1 …………… (ii)
6A + 3B + 2C = O ………. (iii)
From eq(i), we get A = – (B + C)
On putting the value of A in eqs (ii) and (iii), we get
-5{-(B + C)} – 4B – 3C = 1
5B + 5C – 4B – 3C = 1 = B + 2C = 1 ……….. (iv)
and 6 (-(B + C)} + 3B + 2C = 0
⇒ -3B – 4C = 0 ……………. (v)
On solving (iv) & (v), we get C = \(\frac{3}{2}\)
On putting the value of C in eq(iv), we get
B + 2(\(\frac{3}{2}\)) = 1 ⇒ B = -2
Now, put the values of B and C in eq (1) we get
A – 2 + \(\frac{3}{2}\) = 0 ⇒ A = 2 – \(\frac{3}{2}\) = \(\frac{1}{2}\)
∴ A = \(\frac{1}{2}\), B = -2 and C = \(\frac{3}{2}\)
Now, ∫\(\frac{x}{(x-1)(x-2)(x-3)}\) dx [∵ ∫\(\frac{1}{x}\) dx = log|x| + C]
= ∫\(\frac{\mathrm{A}}{(\mathrm{x}-1)}\) dx + ∫\(\frac{\mathrm{B}}{(\mathrm{x}-2)}\) dx + ∫\(\frac{\mathrm{C}}{(\mathrm{x}-3)}\) dx
= \(\frac{1}{2}\) ∫\(\frac{1}{x-1)}\) dx – 2 ∫\(\frac{1}{(x-2)}\) dx + \(\frac{3}{2}\) ∫\(\frac{1}{(x-3)}\) dx
= \(\frac{1}{2}\) log|x – 1| – 2 log|x – 2| + \(\frac{3}{2}\) log|x – 3| + C

Question 3.
\(\frac{2 x}{x^2+3 x+2}\)
Solution:
Let \(\frac{2 x}{x^2+3 x+2}=\frac{2 x}{(x+1)(x+2)}=\frac{A}{(x+1)}+\frac{B}{(x+2)}=\frac{A(x+2)+B(x+1)}{(x+1)(x+2)}\)
⇒ \(\frac{2 \mathrm{x}}{(\mathrm{x}+2)(\mathrm{x}+1)}\) = \(\frac{\mathrm{A}(\mathrm{x}+1)+\mathrm{B}(\mathrm{x}+2)}{(\mathrm{x}+2)(\mathrm{x}+1)}\)
⇒ 2x = x(A + B) + (A + 2B)
On comparing the coefficient of x and constant terms, on both sides, we get
A + B = 2 …………. (1)
and A + 2B = 0 ………….. (ii)
(i) – (ii) ⇒ B = 2 ⇒ B = -2
On putting the value of B in eq (i), we get
A – 2 = 2 ⇒ A = 4
∴ ∫\(\frac{2 \mathrm{x}}{(\mathrm{x}+2)(\mathrm{x}+1)}\) dx = ∫\(\frac{\mathrm{A}}{(\mathrm{x}+2)}\) dx + ∫\(\frac{\mathrm{B}}{(\mathrm{x}+1)}\) dx
= ∫\(\frac{4}{(x+2)}\) dx + ∫\(\frac{(-2)}{(x+1)}\) dx
= 4log |x + 2| – 2log |x + 1| + C

Question 4.
\(\frac{x}{\left(x^2+1\right)(x-1)}\)
Solution:
∫\(\frac{x}{\left(x^2+1\right)(x-1)}\) dx
Let \(\frac{x}{\left(x^2+1\right)(x-1)}\) = \(\frac{A}{x-1}+\frac{B x+C}{x^2+1}\)
⇒ x = A(x² + 1) + (Bx + C) (x – 1) ………….(1)
Substituting x = 1 and O in eq (ii), we get
1 = A(2) and 0 = A – C ⇒ A = \(\frac{1}{2}\) and C = A = \(\frac{1}{2}\)
On equating the coefficient of x² on the both sides in eq (ii), we get
0 = A + B ⇒ B = -A = \(\frac{-1}{2}\)

Question 5.
\(\frac{2}{(1-x)\left(1+x^2\right)}\)
Solution:
Let \(\frac{2}{(1-x)\left(1+x^2\right)}\) = \(\frac{A}{1-x}+\frac{B x+C}{1+x^2}\)
⇒ \(\frac{2}{(1-x)\left(1+x^2\right)}\) = \(\frac{A\left(1+x^2\right)+(B x+C)(1-x)}{(1-x)\left(1+x^2\right)}\)
⇒ A + Ax+ Bx+ C – Bx² – Cx
⇒ 2 = x²(A – B) + x(B – C) + (A + C)
On comparing the coefficients of x², x and constant terms on both sides, we get
A -B = 0 ⇒ A = B ………… (i)
B – C = 0 ⇒ B = C …………. (ii)
and A + C = 2 ………… (iii)
From eq (i) and (ii), we get A = C put this value in eq (iii) we get 2A = 2 ⇒ A = 1
Put the value of A in eq (ii) and (iii), we get B = 1 and C = 1
∴ \(\frac{2}{(1-x)\left(1+x^2\right)}=\frac{1}{1-x}+\frac{x+1}{1+x^2}\)
= ∫\(\frac{1}{1-x}\) dx + \(\frac{1}{2}\) ∫\(\frac{2 x}{x^2+1}\) dx + ∫\(\frac{1}{x^2+1}\) dx
Let x² + 1 = t ⇒ 2x dx = dt
∫\(\frac{2 x}{x^2+1}\) dx = ∫\(\frac{1}{t}\) dt = log t
= log |1 – x| + \(\frac{1}{2}\) log(1 + x²) + tan^-1x +

Question 6.
\(\frac{3 x-1}{(x+2)^2}\)
Solution:
Let \(\frac{3 x-1}{(x+2)^2}\) = \(\frac{A}{x+2}+\frac{B}{(x+2)^2}\)
⇒ 3x – 1 = A(x + 2) + B
On equating the coefficient of x and constant terms on both sides, we get A = 3 and
2A + B = -1 ⇒ 2(3) + B = -1 ⇒ B = -7
∴ \(\frac{3 x-1}{(x+2)^2}\) = \(\frac{3}{x+2}+\frac{7}{(x+2)^2}\)
∴ ∫\(\frac{3 x-1}{(x+2)^2}\) = 3 ∫\(\frac{1}{(x+2)}\) dx – 7 ∫\(\frac{1}{(x+2)^2}\) dx
= 3 log |x + 2| – 7 \(\left(\frac{-1}{x+2}\right)\) + C
= 3 log |x + 2| + \(\frac{7}{x+2}\) + C

Question 7.
\(\frac{1}{x\left(x^n+1\right)}\)
[Hint: Multiply numerator and denominator by x^n-1 and put xⁿ = t]
Solution:
Let I = ∫\(\frac{1}{x\left(x^n+1\right)}\) dx
Put xⁿ = t ⇒ nx^n-1dx = dt ⇒ x^n-1 dx = \(\frac{1}{n}\) dt
∴ ∫\(\frac{\mathrm{x}^{\mathrm{n}-1}}{\mathrm{x}^{\mathrm{n}}\left(\mathrm{x}^{\mathrm{n}}+1\right)}\) dx = \(\frac{1}{n} \int \frac{1}{t(t+1)} d t\) ………….. (i)
Now, \(\frac{1}{t(t+1)}=\frac{A}{t}+\frac{B}{(t+1)}\)
1 = A(1 + t) + Bt ………….. (ii)
On substituting t = 0, -1 in equation (ii), we get
A = 1 and B = -1
∴ \(\frac{1}{t(t+1)}=\frac{1}{t}-\frac{1}{(t+1)}\)
∴ \(\frac{1}{n} \int\left[\frac{1}{t}-\frac{1}{(t+1)}\right]\) dt = \(\frac{1}{n}\) [log |t| – log |t + 1| + C
= \(\frac{1}{n}\) [log |xⁿ| = log(xⁿ + 1)] + C [put t = xⁿ]
= \(\frac{1}{n} \log \left|\frac{x^n}{x^n+1}\right|\) + C

Question 8.
\(\frac{1}{\left(e^x-1\right)}\) [Hint: Put e^x = t]
Solution:
Let I = ∫\(\frac{1}{\left(e^x-1\right)}\) dx
On mutliplying numerator & denominator by e^-x,
I = ∫\(\frac{e^{-x}}{1-e^{-x}}\) dx
Put 1 – e^-x = t ⇒ -e^-x (-1) = \(\frac{d t}{d x}\) ⇒ e^-x dx = dt
∴ I = ∫\(\frac{d t}{t}\) = log |t| + C = log |1 – e^-x| + C
= \(\log \left|\frac{e^x-1}{e^x}\right|\) + C

Question 9.
\(\frac{e^x}{\left(1+e^x\right)\left(2+e^x\right)}\)
Solution:
Let I = ∫\(\frac{e^x}{\left(1+e^x\right)\left(2+e^x\right)}\) dx
Put e^x = t ⇒ e^x dx = dt
∴ I = ∫\(\frac{e^x}{1+t)(2+t)} \frac{d t}{e^x}\) = ∫\(\frac{1}{(1+t)(2+t)}\) dt
Let \(\frac{1}{(1+t)(2+t)}\) = \(\frac{A}{(1+t)}+\frac{B}{(2+t)}\) = \(\frac{A(2+t)+B(1+t)}{(1+t)(2+t)}\)
⇒ 1 = (A + B)t + (2A + B)
On equating the coefficients of t and constant terms on both sides, we get A + B = O and 2A + B = 1
On solving both equations we get A = 1 & B = -1
I = ∫\(\frac{1}{(1+t)}\) dt – ∫\(\frac{1}{(2+t)}\) dt
= log |1 + t| – log |2 + t| + C
= \(\log \left|\frac{1+t}{2+t}\right|\) + C
= \(\log \left|\frac{1+e^x}{2+e^x}\right|\) + C

Question 10.
\(\frac{1}{\left(x^2+1\right)\left(x^2+4\right)}\)
Solution:
∫\(\frac{1}{\left(x^2+1\right)\left(x^2+4\right)}\) dx
Let \(\frac{1}{\left(x^2+1\right)\left(x^2+4\right)}\) = \(\frac{A x+B}{x^2+1}\) + \(\frac{C x+D}{x^2+1}\)
⇒ 1 = (Ax + B) (x² + 4) + (Cx + D) (x² + 1)
On comparing the coefficients of x³, x², x and constant terms on both sides, we get
A + C = 0, B + D = 0, 4A + C = 0 and 4B + D = 1
On solving these equations, we get
A = 0, C = OB = and D = \(\frac{-1}{3}\)
∴ ∫\(\frac{1}{\left(x^2+1\right)\left(x^2+4\right)}\) dx = \(\frac{1}{3}\left(\int \frac{1}{\left(x^2+1\right)}-\frac{1}{\left(x^2+4\right)}\right)\) dx
= \(\frac{1}{3}\left[\tan ^{-1} x-\frac{1}{2} \tan ^{-1}\left(\frac{x}{2}\right)\right]\) + C

III. Integrate the Following Rational Functions.

Question 1.
\(\frac{x}{(x-1)^2(x+2)}\)
Solution:
Let ∫\(\frac{\mathrm{x}}{(\mathrm{x}-1)^2(\mathrm{x}+2)}=\frac{\mathrm{A}}{\mathrm{x}-1}+\frac{\mathrm{B}}{(\mathrm{x}-1)^2}+\frac{\mathrm{C}}{\mathrm{x}+2}\) ………… (1)
x = A(x – 1)(x + 2) + B(x + 2) + C(x – 1)² ………….. (2)
Put x = 1, -2 in eq (2) we get
1 = B(1 + 2) ⇒ B = 1/3
-2 = C(-2 – 1)² ⇒ C = \(\frac{2}{9}\)
On comparing coefficient of x² in eqn (2) then
0 = A + C ⇒ A = -C = \(\frac{2}{9}\) ⇒ A = \(\frac{2}{9}\)
Substituting A, B & C values in eqn (1) then

Question 2.
\(\frac{3 x+5}{x^3-x^2-x+1}\)
Solution:
x³ – x² – x + 1 = (x – 1)²(x + 1)
Let \(\frac{3 x+5}{x^3-x^2-x+1}=\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C}{x+1}\) …………. (1)
3x + 5 = A(x² – 1) + B(x + 1) + C(x – 1)²
3x + 5 = x²(A + C) + x (B – 2C) + (B + C – A) ……….. (2)
Put x = 1 then 8 = 2B ⇒ B = 4
Put x = -1 then 2 = 4C ⇒ C = 1/2
On comparing coefficients of x² in eqn (2) then
A + C = O ⇒ A = -C = -1/2
Substitute A, B, C values in eqn(1) then

Question 3.
\(\frac{2 x-3}{\left(x^2-1\right)(2 x+3)}\)
Solution:
Let \(\frac{2 x-3}{\left(x^2-1\right)(2 x+3)}\) = \(\frac{A}{x-1}+\frac{B}{x+1}+\frac{C}{2 x+3}\) ………… (1)
2x – 3 = A(2x + 3)(x + 1) + B(x – 1)(2x + 3) + C(x – 1)(x + 1) ………… (2)
Put x = 1 in eqn (2) then -1 = 10A ⇒ A = \(\frac{-1}{10}\)
Put x = -1 in eqn(2) then -5 = B(-2) ⇒ B = \(\frac{5}{2}\)
Put x = \(\frac{-3}{2}\) in eqn(2) then -6 = C(\(\frac{-3}{2}\) -1)(\(\frac{-3}{2}\) + 1)
-6 = C(\(\frac{-5}{2}\))(\(\frac{-1}{2}\)) ⇒ -24 = 5C ⇒ C = –\(\frac{24}{5}\)
Substitute A, B, C values in eqn(1) then

Question 4.
\(\frac{5 x}{(x+1)\left(x^2-4\right)}\)
Solution:
Let \(\frac{5 x}{(x+1)\left(x^2-4\right)}=\frac{A}{(x+1)}+\frac{B}{(x+2)}+\frac{C}{(x-2)}\) …….. (1)
5x = A(x² – 4) + B(x + 1)(x – 2) + C(x + 1)(x + 2) ………… (2)
Put x = -1 in eqn(2) then -5 ⇒ -3A = -5 ⇒ A = \(\frac{5}{3}\)
Put x = 2 in eqn(2) then 10 = C(3)(4) ⇒ C = \(\frac{5}{6}\)
Put x = -2 in eqn(2) then -10 = B(-4)(-1) ⇒ B = –\(\frac{5}{2}\)
Substitute A, B, C values in eqn(1) then

Question 5.
\(\frac{x^3+x+1}{x^2-1}\)
Solution:
Let \(\frac{x^3+x+1}{x^2-1}\) = x + \(\frac{2 x+1}{x^2-1}\) ………….. (1)
Consider \(\frac{2 x+1}{x^2-1}=\frac{A}{(x+1)}+\frac{B}{(x-1)}\) ……….. (2)
2x + 1 = A(x + 1) + B(x – 1) …………. (3)
Put x = 1 in eqn(3) then 3 = 2A ⇒ A = \(\frac{3}{2}\)
Put x = -1 in eqn(3) then -1 = -2B ⇒ D = \(\frac{1}{2}\)
Substitute A, B values in eq(2) then
\(\frac{2 x+1}{x^2-1}\) = \(\frac{3 / 2}{x-1}+\frac{1 / 2}{x+1}\) ……….. (4)
From 1 & 4 we get
\(\frac{x^3+x+1}{x^2-1}\) = x + \(\frac{3}{2(x-1)}\) + \(\frac{1}{2(x+1)}\)
∴ ∫\(\frac{x^3+x+1}{x^2-1}\) dx = v x dx + \(\frac{3}{2}\) ∫\(\frac{1}{x-1}\) dx + \(\frac{1}{2}\) ∫\(\frac{1}{x+1}\) dx
= \(\frac{x^2}{2}\) + \(\frac{3}{2}\) log|x – 1| + \(\frac{1}{2}\)log|x + 1| + C

Question 6.
\(\frac{1}{x^4-1}\)
Solution:
Consider \(\frac{1}{x^4-1}\) = \(\frac{1}{(x-1)(x+1)\left(x^2+1\right)}\)
Let \(\frac{1}{(x+1)(x-1)\left(x^2+1\right)}=\frac{A}{(x+1)}+\frac{B}{(x-1)}+\frac{C x+D}{\left(x^2+1\right)}\) ………… (1)
⇒ 1 = A(x + 1) (x² + 1) + B(x – 1) (x² + 1) + (Cx + D)(x² – 1)
Put x = -1 then 1 = -4B ⇒ B=—1/4
Put x = 1 then 1 = 4A ⇒ A = 1/4
Put x = 0 then 1 = A – B – D
D = \(\frac{1}{4}\) + \(\frac{1}{4}\) – 1 = -2/4 = -1/2
On comparing the coefficients of x on both sides,
0 = A + B – C ⇒ C = A + B = \(\frac{1}{4}\) + \(\frac{1}{4}\) = 0
Substituting the values of A, B, C & D in (1),
\(\frac{1}{(x-1)(x+1)\left(x^2+1\right)}\) = \(\frac{1 / 4}{(x-1)}+\frac{-1 / 4}{x+1}+\frac{-1 / 2}{x^2+1}\)
∴ ∫\(\frac{1}{x^4-1}\) dx = \(\frac{1}{4}\) ∫\(\frac{1}{x-1}\) – \(\frac{1}{4}\) ∫\(\frac{1}{x+1}\) dx – \(\frac{1}{2}\) ∫\(\frac{1}{x^2+1}\) dx
= \(\frac{1}{4}\) log |x – 1| – \(\frac{1}{4}\) log|x + 1| – \(\frac{1}{2}\) tan^-1 + C
= \(\frac{1}{4}\) log\(\frac{x-1}{x+1}\) – \(\frac{1}{2}\)tan^-1x + C [∵ ∫\(\frac{1}{x^2+1}\) dx = tan^-1x + C]

Question 7.
\(\frac{\cos x}{(1-\sin x)(2-\sin x)}\) [Hint: Put sin x = t]
Solution:
∫\(\frac{\cos x}{(1-\sin x)(2-\sin x)}\) dx
Put sin x = t ⇒ cos xdx = dt
Consider \(\frac{1}{(1-t)(2-t)}\) = \(\frac{A}{1-t}+\frac{B}{2-t}\) ………… (1)
1 = A(2 – t) + B(1 – t)
Put t = 1, 2 then A = 1 & B = -1
Substituting A, B values in eq(1) then

Question 8.
\(\frac{\left(x^2+1\right)\left(x^2+2\right)}{\left(x^2+3\right)\left(x^2+4\right)}\)
Solution:

Question 9.
\(\frac{2 x}{\left(x^2+1\right)\left(x^2+3\right)}\)
Solution:
Consider ∫\(\frac{2 x}{\left(x^2+1\right)\left(x^2+3\right)}\) dx
Put x² = t ⇒ 2x dx = dt
Let \(\frac{1}{(t+1)(t+3)}=\frac{A}{t+1}+\frac{B}{t+3}=\frac{A(t+3)+B(t+1)}{(t+1)(t+3)}\)
⇒ 1 = A(t + 3) + B(t + 1)
Put t = -1 then 2A = 1 ⇒ A = 1/2
Put t = -3 then -2B = 1 ⇒ B = -1/2
∴ ∫\(\frac{\mathrm{dt}}{(\mathrm{t}+1)(\mathrm{t}+3)}\) = \(\int \frac{\frac{1}{2}}{t+1} d t-\int \frac{1 / 2}{t+3} d t\)
= \(\frac{1}{2}\) ∫\(\frac{1}{t+1}\) dt – \(\frac{1}{2}\) ∫\(\frac{1}{t+3}\) dt
= \(\frac{1}{2}\) log|t + 1| – \(\frac{1}{2}\) log|t + 3| + C
= \(\frac{1}{2}\) log|x² + 1| – \(\frac{1}{2}\) log|x² + 3| + C
= \(\frac{1}{2}\) log\(\left|\frac{x^2+1}{x^2+3}\right|\) + C

Question 10.
\(\frac{1}{x\left(x^4-1\right)}\)
Solution:

Question 11.
\(\frac{1}{x-x^3}\)
Solution:
\(\int \frac{1}{x-x^3} d x\) = \(\int \frac{1}{x\left(1-x^2\right)} d x=\int \frac{1}{x(1-x)(1+x)} d x\)
Consider \(\frac{1}{x(1-x)(1+x)}=\frac{A}{x}+\frac{B}{1-x}+\frac{C}{1+x}\) ……….. (1)
1 = A(1 – x²) + Bx(1 + x) + Cx(1 – x) ………….. (2)
Put x = 0 in eq (2) then 1 = A
Put x = 1 in eq (2) then 1 = 2B ⇒ B = 1/2
Put x = -1 in eq (2) then 1 = -2C ⇒ C = -1/2
Substitute A, B, C values in eqn(1) then
\(\frac{1}{x\left(1-x^2\right)}\) = \(\frac{1}{x}+\frac{1}{2(1-x)}-\frac{1}{2(1+x)}\)
∫\(\frac{1}{x\left(1-x^2\right)}\) dx = ∫\(\frac{1}{x}\) dx + \(\frac{1}{2}\) ∫\(\frac{1}{1-x}\) dx – \(\frac{1}{2}\) ∫\(\frac{1}{1+x}\) dx
= log x – \(\frac{1}{2}\) log|1 – x| – \(\frac{1}{2}\) log|1 + x| + C
= log x² – \(\frac{1}{2}\) [log(1 – x²)] + C
= \(\frac{1}{2}\) log \(\frac{x^2}{1-x^2}\) + C

Question 12.
\(\frac{1}{x^{\frac{1}{2}}+x^{\frac{1}{3}}}\) [Hint: \(\frac{1}{x^{\frac{1}{3}}\left(1+x^{\frac{1}{6}}\right)}\), put x = t⁶]
Solution:

Question 13.
\(\frac{5 x}{(x+1)\left(x^2+9\right)}\)
Solution:
∫\(\frac{5 x}{(x+1)\left(x^2+9\right)}\) dx
Consider \(\frac{5 x}{(x+1)\left(x^2+9\right)}\) = \(\frac{A}{x+1}+\frac{B x+C}{x^2+9}\) ………… (1)
5x = A(x² + 9) + (Bx + C) (x + 1) ……….. (2)
Put x = -1 in eqn(2) then 5(-1) ⇒ 10A = -5 ⇒ A = -1/2
Comparing the coefficient of x² on both sides,
0 = A + B ⇒ B = -A = 1/2
Comparing the coefficient of x² on both sides,
5 = B + C ⇒ 5 = \(-\frac{1}{2}\) + C ⇒ C = 5 – \(-\frac{1}{2}\) = \(-\frac{9}{2}\)
Substituting A, B, C values in eqn (1) then

Question 14.
\(\frac{x^2+x+1}{(x+1)^2(x+2)}\)
Solution:
∫\(\frac{x^2+x+1}{(x+1)^2(x+2)}\) dx
Consider \(\frac{x^2+x+1}{(x+1)^2(x+2)}=\frac{A}{(x+1)}+\frac{B}{(x+1)^2}+\frac{C}{(x+2)}\) ……….. (1)
x² + x + 1 = A(x + 1) (x + 2) + B(x + 2) + C(x + 1)² ………….. (2)
Put x = -1 in eq(2) then 1 – 1 + 1 = B ⇒ B = 1
Put x = -2 in eq(2) then 4 – 2 + 1 = C(-1)² ⇒ C = 3
Comparing the coefficients of x² on both sides,
1 = A + C ⇒ A = 1 – C = 1 – 3 = -2
Substituting A, B, C values in eq (1) then
\(\frac{x^2+x+1}{(x+1)^2(x+2)}\) = \(\frac{-2}{x+1}+\frac{1}{(x+1)^2}+\frac{3}{x+2}\)
∫\(\frac{x^2+x+1}{(x+1)^2(x+2)}\) dx
= – ∫\(\frac{1}{x+1}\) dx + ∫\(\frac{1}{(x+1)^2}\) dx + 3∫\(\frac{1}{x+2}\) dx
= -2 log|x + 1| – \(\frac{1}{x+1}\) + 3 log|x + 2| + C

Practicing the AP Inter 2nd Year Maths Study Material and AP Inter 2nd Year Maths Chapter 7 Integrals Solutions Exercise 7d will help students to clear their doubts quickly.

Inter 2nd Year Maths Integrals Solutions Exercise 7d

Integrals Exercise 7d Solutions

I. Integrate the functions

Question 1.
\(\frac{3 x^2}{x^6+1}\)
Solution:
Consider \(\frac{3 x^2}{x^6+1}\)
Let x³ = t ⇒ 3x² dx = dt
∴ ∫ \(\frac{3 x^2}{x^6+1}\) dx = ∫ \(\frac{\mathrm{dt}}{\mathrm{t}^2+1}\)
= tan^-1t + C = tan^-1 (x³) + C [∵ ∫ \(\frac{1}{x^2+a^2}\) dx = \(\frac{1}{a}\) tan^-1 \(\frac{x}{a}\) + C]

Question 2.
\(\frac{1}{\sqrt{1+4 x^2}}\)
Solution:
Consider \(\frac{1}{\sqrt{1+4 x^2}}\)
Let 2x = t ⇒ 2dx = dt

Question 3.
\(\frac{1}{\sqrt{(2-x)^2+1}}\)
Solution:

Question 4.
\(\frac{1}{\sqrt{9-25 x^2}}\)
Solution:

Question 5.
\(\frac{3 x}{1+2 x^4}\)
Solution:

Question 6.
\(\frac{x^2}{1-x^6}\)
Solution:
Consider \(\frac{x^2}{1-x^6}\)
Let x³ = t ⇒ 3x² dx = dt
∴ \(\int \frac{x^2}{1-x^6} d x=\frac{1}{3} \int \frac{d t}{1-t^2}\) [∵ \(\frac{d x}{a^2-x^2}=\frac{1}{2 a} \log \left|\frac{a+x}{a-x}\right|\) + C]
= \(\frac{1}{3}\left[\frac{1}{2} \log \left|\frac{1+t}{1-t}\right|\right]\) + C
= \(\frac{1}{6} \log \left|\frac{1+x^3}{1-x^3}\right|\) + C

Question 7.
\(\frac{x-1}{\sqrt{x^2-1}}\)
Solution:

Question 8.
\(\frac{x^2}{\sqrt{x^6+a^6}}\)
Solution:
Consider \(\frac{x^2}{\sqrt{x^6+a^6}}\)
Let x³ = t ⇒ 3x² dx = dt
∴ \(\frac{x^2}{\sqrt{x^6+a^6}}\) dx = \(\frac{1}{3} \int \frac{\mathrm{dt}}{\sqrt{\mathrm{t}^2+\left(\mathrm{a}^3\right)^2}}\)
= \(\frac{1}{3}\) log |t + \(\sqrt{t^2+\left(a^3\right)^2}\)| + C
[∵ \(\frac{\mathrm{dx}}{\sqrt{\mathrm{x}^2+\mathrm{a}^2}}=\log \left|\mathrm{x}+\sqrt{\mathrm{x}^2+\mathrm{a}^2}\right|\) + C]
= \(\frac{1}{3}\) log |x³ + \(\sqrt{x^6+a^6}\)| + C

Question 9.
\(\frac{\sec ^2 x}{\sqrt{\tan ^2 x+4}}\)
Solution:
Consider \(\frac{\sec ^2 x}{\sqrt{\tan ^2 x+4}}\)
Let tan x = t ⇒ sec² dx = dt
∴ \(\int \frac{\sec ^2 x}{\sqrt{\tan ^2 x+4}}\) dx
= latex]\int \frac{d t}{\sqrt{t^2+2^2}}[/latex]
= \(\log \left|t+\sqrt{t^2+4}\right|\) + C
[∵ \(\frac{\mathrm{dx}}{\sqrt{\mathrm{x}^2+\mathrm{a}^2}}=\log \left|\mathrm{x}+\sqrt{\mathrm{x}^2+\mathrm{a}^2}\right|\) + C]
= \(\log \left|\tan x+\sqrt{\tan ^2 x+4}\right|\) + C

Question 10.
\(\frac{\cos x}{\sqrt{4-\cos ^4 x}}\)
Solution:
Consider \(\frac{\cos x}{\sqrt{4-\cos ^4 x}}\)
= \(\frac{\cos x}{\sqrt{4-\left(\cos ^2 x\right)^2}}=\frac{\cos x}{\sqrt{4-\left(1-\sin ^2 x\right)^2}}\)
Let sin x = t ⇒ cos x dx
∴ I = \(\int \frac{\cos x}{\sqrt{4-\cos ^4 x}}\) dx
= \(\int \frac{d t}{\sqrt{2^2-(t)^2}}\)
Using standard result
I = \(\frac{1}{\sqrt{3}} \sin ^{-1}\left(\frac{t}{\sqrt{3}}\right)\) + C
Substitute t = sin x = \(\frac{1}{\sqrt{3}} \sin ^{-1}\left(\frac{t}{\sqrt{3}}\right)\) + C

II. Integrate the following functions

Question 1.
\(\frac{1}{\sqrt{x^2+2 x+2}}\)
Solution:

Question 2.
\(\frac{1}{9 x^2+6 x+5}\)
Solution:

Question 3.
\(\frac{1}{\sqrt{7-6 x-x^2}}\)
Solution:
Consider \(\frac{1}{\sqrt{7-6 x-x^2}}\)
7 – 6x – x² can be written as 7 – (x² + 6x + 9 – 9)
7 – (x² + 6x + 9 – 9) = 16 – (x² + 6x + 9)
= 16 – (x + 3)² = 4² – (x + 3)²
∴ \(\int \frac{1}{\sqrt{7-6 x-x^2}}\) dx
= \(\int \frac{1}{\sqrt{4^2-(x+3)^2}}\) dx
Let x + 3 = t ⇒ dx = dt
∴ \(\int \frac{1}{\sqrt{4^2-(x+3)^2}}\) dx
= \(\int \frac{1}{\sqrt{4^2-t^2}}\) dt
= \(\sin ^{-1}\left(\frac{t}{4}\right)\) + C [∵ \(\int \frac{d x}{\sqrt{x^2-a^2}}=\sin ^{-1}\left(\frac{x}{a}\right)\) + C]
= \(\sin ^{-1}\left(\frac{x+3}{4}\right)\) + C

Question 4.
\(\frac{1}{\sqrt{(x-1)(x-2)}}\)
Solution:
We have (x – 1)(x – 1) = x² – 3x + 2

Question 5.
Find the integral of \(\frac{1}{\sqrt{8+3 x-x^2}}\)
Solution:

Question 6.
\(\frac{1}{\sqrt{(x-a)(x-b)}}\)
Solution:
Consider \(\frac{1}{\sqrt{(x-a)(x-b)}}\)
(x – a)(x – b) can be written as x² – (a + b)x + ab
∴ x² – (a + b)x + ab
= x² – (a + b)x + \(\frac{(a+b)^2}{4}-\frac{(a+b)^2}{4}+ab\)
= \(\left[x-\left(\frac{a+b}{2}\right)\right]^2-\frac{(a-b)^2}{4}\)

Question 7.
\(\frac{x+2}{\sqrt{x^2-1}}\)
Solution:
∫ \(\frac{x+2}{\sqrt{x^2-1}}\) dx
= ∫ \(\frac{x}{\sqrt{x^2-1}}\) dx + ∫ \(\frac{2}{\sqrt{x^2-1}}\) dx ………….. (i)
Now, I₁ = ∫ \(\frac{x}{\sqrt{x^2-1}}\) dx
Let, x² – 1 = t ⇒ 2x dx = dt ⇒ dx = \(\frac{d t}{2 x}\)
I₁ = \(\int \frac{\mathrm{x}}{\sqrt{\mathrm{t}}} \times \frac{\mathrm{dt}}{2 \mathrm{x}}=\frac{1}{2} \int \frac{\mathrm{dt}}{\sqrt{\mathrm{t}}}=\frac{1}{2} \int \mathrm{t}^{-1 / 2} \mathrm{dt}\)
= \(\frac{1}{2}\) [2t^1/2] = \(\sqrt{t}\)
= \({\sqrt{x^2-1}}\) + C₁ [∵ t = x² – 1]
Now, I₂ = 2 ∫ \(\frac{1}{\sqrt{x^2-1}}\) dx = 2 log|x + \({\sqrt{x^2-1}}\)| + C₂
[∵ \(\int \frac{d x}{\sqrt{x^2-a^2}}=\sin ^{-1}\left(\frac{x}{a}\right)\) + C]
On Putting the values of I₁ & I₂ in eq (i) we get
I = \({\sqrt{x^2-1}}\) + 2 log|x + \({\sqrt{x^2-1}}\)| + C
Where, C = C₁ + C₂

Question 8.
\(\frac{4 x+1}{\sqrt{2 x^2+x-3}}\)
Solution:
Consider \(\frac{4 x+1}{\sqrt{2 x^2+x-3}}\)
Let 2x² + x – 3 = t
⇒ (4x + 1)dx = dt ⇒ dx = \(\frac{d t}{4 x+1}\)
∴ \(\int \frac{4 x+1}{\sqrt{2 x^2+x-3}}\) dx = \(\int \frac{4 x+1}{\sqrt{t}} \times \frac{d t}{4 x+1}\)
= \(\int \frac{1}{\sqrt{t}}\) dt
= 2\( \sqrt{t}\) + C = 2\(\sqrt{2 x^2+x-3}\) + C [∵ t = 2x² + x – 3]

III. Integrate the following functions

Question 1.
\(\frac{5 x-2}{1+2 x+3 x^2}\)
Solution:
∫\(\frac{5 x-2}{1+2 x+3 x^2}\) dx
Let 5x – 2 = A\(\frac{d}{d x}\)(1 + 2x + 3x²) + B
⇒ 5x – 2 = A(2 + 6x) + B ⇒ 5x – 2 = 6Ax + (2A + B)
Equating the coefficients of x and constant term on both sides, we get
5 = 6A ⇒ A = \(\frac{5}{6}\) and
2A + B = -2 ⇒ \(\frac{5}{3}\) + B = -2 ⇒ B = \(-\frac{11}{3}\)
∴ 5x – 2 = \(\frac{5}{6}\)(2 + 6x) + (\(-\frac{11}{3}\))
∴ \(\int \frac{5 x-2}{1+2 x+3 x^2} d x=\int \frac{\frac{5}{6}(2+6 x)-\frac{11}{3}}{1+2 x+3 x^2} d x\)
= \(\frac{5}{6} \int \frac{2+6 x}{1+2 x+3 x^2} d x-\frac{11}{3} \int \frac{1}{1+2 x+3 x^2} d x\)
Let I₁ = \(\int \frac{2+6 x}{1+2 x+3 x^2}\)dx and I₂ = \(\int \frac{1}{1+2 x+3 x^2}\)dx
I = \(\frac{5}{6}\)I₁ + (-\(\frac{11}{3}\))I₂ …………. (1)
Now, I₁ = \(\int \frac{2+6 x}{1+2 x+3 x^2}\)dx
Let 1 + 2x + 3x² = t ⇒ (2+ 6x)dx = dt
∴ I₁ = \(\int \frac{d t}{t}\) = log|t| + C₁
⇒ I₁ = log(1 + 2x + 3x²) + C₁ ……(2)
Also I₂ = \(\int \frac{1}{1+2 x+3 x^2} d x\)
1 + 2x + 3x² can be written as

Question 2.
\(\frac{6 x+7}{\sqrt{(x-5)(x-4)}}\)
Solution:
∫ \(\frac{6 x+7}{\sqrt{(x-5)(x-4)}}\) dx = ∫ \(\frac{6 x+7}{\sqrt{x^2-9 x+20}}\) dx
Let 6x + 7 = A\(\frac{d}{d x}\)(x² – 9x + 20) + B
⇒ 6x + 7 = A(2x – 9) + B
⇒ 6x + 7 = 2Ax + (-9A + B)
On equating the coefficients of x and constant term on both sides, we get
2A = 6 ⇒ A = 3; -9A + B = 7 ⇒ B = 34

Let x² – 9x + 20 = t ⇒ (2x – 9)dx = dt
∴ I₁ = \(\int \frac{d t}{\sqrt{t}}[latex] = 2[latex]\sqrt{t}\) + C₁ = 2\(\sqrt{x^2-9 x+20}\) + C₁ ……………..(2)
∴ I₂ = \(\int \frac{1}{\sqrt{x^2-9 x+20}}\) dx
x² – 9x + 20 can be written as
x² – 9x + 20 + \(\frac{81}{4}-\frac{81}{4}\)
∴ x² – 9x + 20 + \(\frac{81}{4}-\frac{81}{4}\)
= \(\left(x-\frac{9}{2}\right)^2-\frac{1}{4}=\left(x-\frac{9}{2}\right)^2-\left(\frac{1}{2}\right)^2\)
∴ I₂ = \(\int \frac{1}{\sqrt{\left(x-\frac{9}{2}\right)^2-\left(\frac{1}{2}\right)^2}}\) dx
= log\(\left|\left(x-\frac{9}{2}\right)+\sqrt{\left(x-\frac{9}{2}\right)^2-\left(\frac{1}{4}\right)}\right|\) + C₂
[∵ \(\int \frac{\mathrm{dx}}{\sqrt{\mathrm{x}^2-\mathrm{a}^2}}=\log \left|\mathrm{x}+\sqrt{\mathrm{x}^2-\mathrm{a}^2}\right|\)
= log \(\left(x-\frac{9}{2}\right)+\sqrt{x^2-9 x+20}\) + C₂ ……………………… (3)
On Substituting the values of I₁ & I₂ for equations (2) and (3) in (1), we get
\(\int \frac{6 x+7}{\sqrt{x^2-9 x+20}}\) dx
= \(3\left[2 \sqrt{x^2-9 x+20}\right]+34 \log \left[\left(x-\frac{9}{2}\right)+\sqrt{x^2-9 x+20}\right]\) + C
= \(6 \sqrt{x^2-9 x+20}+34 \log \left[\left(x-\frac{9}{2}\right)+\sqrt{x^2-9 x+20}\right]\) + C

Question 3.
\(\frac{x+2}{\sqrt{4 x-x^2}}\)
Solution:
Let I = ∫ \(\frac{x+2}{\sqrt{4 x-x^2}}\) dx
Let x + 2 = A\(\frac{d}{d x}\)(4x – x²) + B ⇒ x + 2 = A(4 – 2x) + B
On equating the coefficients of x and constant term on both sides, we get
-2A = 1 ⇒ A = \(-\frac{1}{2}\); 4A + B = 2 ⇒ B = 4
⇒ (x + 2) = \(-\frac{1}{2}\)(4 – 2x) + 4
∴ I = \(\int \frac{x+2}{\sqrt{4 x-x^2}} d x=\int \frac{-\frac{1}{2}(4-2 x)+4}{\sqrt{\left(4 x-x^2\right)}} d x\)
= \(-\frac{1}{2} \int \frac{(4-2 x)}{\sqrt{\left(4 x-x^2\right)}} d x+4 \int \frac{1}{\sqrt{\left(4 x-x^2\right)}} d x\)
let I₁ = \(\int \frac{4-2 x}{\sqrt{4 x-x^2}} d x\) and I₂ = \(\int \frac{1}{\sqrt{4 x-x^2}} d x\)
∴ I = –\(\frac{1}{2}\)I₁ + 4I₂
Now, I₁ = \(\int \frac{4-2 x}{\sqrt{4 x-x^2}} d x\) [∵ \(\int \frac{1}{\sqrt{x}} d x=2 \sqrt{x}+C\)]
Let 4x – x² = t ⇒ (4 – 2x)dx = dt
∴ I₁ = \(\int \frac{\mathrm{dt}}{\sqrt{\mathrm{t}}}=2 \sqrt{\mathrm{t}}\) + C₁ = \(2 \sqrt{4 \mathrm{x}-\mathrm{x}^2}\) + C₁ …….(2)
Now I₂ = \(\int \frac{1}{\sqrt{4 x-x^2}} d x\)
⇒ 4x – x² = -(-4x + x²)
= -{(x – 2)² – 4} = (2)² – (x – 2)²
∴ I₂ = \(\int \frac{1}{\sqrt{(2)^2-(x-2)^2}} d x=\sin ^{-1}\left(\frac{x-2}{2}\right)\) …..(3)
Substituting (2) and (3) in (1), we get
\(\int \frac{x+2}{\sqrt{4 x-x^2}} d x=-\frac{1}{2}\left(2 \sqrt{4 x-x^2}\right)+4 \sin ^{-1}\left(\frac{x-2}{2}\right)\) + C₂ ……….. (3)
[∵ \(\int \frac{1}{\sqrt{a^2-x^2}} d x=\sin ^{-1}\left(\frac{x}{a}\right)\)]
On substituting the values of I₁ & I₂ from eq (2) & (3) in eq (1), we get
∴ \(\int \frac{x+2}{\sqrt{4 x-x^2}} d x=-\frac{1}{2}\left[2 \sqrt{4 x-x^2}\right]+4 \sin ^{-1}\left(\frac{x-2}{2}\right)\) + C
= \(-\sqrt{4 x-x^2}+4 \sin ^{-1}\left(\frac{x-2}{2}\right)\) + C [∵ –\(\frac{1}{2}\)C₁ + 4C₂ = C]

Question 4.
\(\frac{x+2}{\sqrt{x^2+2 x+3}}\)
Solution:
∫ \(\frac{x+2}{\sqrt{x^2+2 x+3}}\) dx
Let x + 2 = A\(\frac{d}{d x}\)(x² + 2x + 3) + B
⇒ x +2 = A(2x + 2) + B
⇒ x + 2 = 2Ax + (2A + B)
On equating the coefficient of x & constant term on both sides, we get
2A = 1 ⇒ A = \(\frac{1}{2}\) &
2A + B = 2 ⇒ 2 × \(\frac{1}{2}\) +B = 2 ⇒ B = 2 – 1 = 1
∴ (x + 2) = \(\frac{1}{2}\) (2x + 2) + 1

= log|x + 1 + \(\sqrt{(x+1)^2+2}\)| + C₂
= log|x + 1 + \(\sqrt{x^2+2 x+3}\)| + C₂ …………. (3)
On putting the values of I₁ & I₂ from eq(2) & (3) in eq (1), we get
∫ \(\frac{x+2}{\sqrt{x^2+2 x+3}}\) dx
= \(\frac{1}{2}\left[2 \sqrt{x^2+2 x+3}\right]+\log \left|(x+1)+\sqrt{x^2+2 x+3}\right|\) + C
[∵ \(\frac{1}{2}\)C₁ + 4C₂ = C]
= \(\sqrt{x^2+2 x+3}+\log \mid(x+1)+\sqrt{x^2+2 x+3}\) + C

Question 5.
\(\frac{x+3}{x^2-2 x-5}\)
Solution:
∫ \(\frac{x+3}{x^2-2 x-5}\) dx
Let (x + 3) = A\(\frac{d}{d x}\)(x² – 2x – 5) + B
⇒ (x + 3) = A(2x – 2) + B
⇒ x + 3 = 2Ax – 2A + B
On equating the coefficients of x and constant term on both sides, we get
2A = 1 ⇒ A = \(\frac{1}{2}\)
-2A + B = 3 ⇒ B = 4
⇒ (x + 3) = \(\frac{1}{2}\)(2x – 2) + 4

On substituting the values of I₂ & I₂ from eq(2) and (3) in (1), we get
∫ \(\frac{x+3}{x^2-2 x-5}\) dx
= \(\frac{1}{2} \log \left|x^2-2 x-5\right|+4\left[\frac{1}{2 \sqrt{6}} \log \left|\frac{x-1-\sqrt{6}}{x-1+\sqrt{6}}\right|\right]\)
[∵ \(\frac{1}{2}\)C₁ + 4C₂ = C]
= \(=\frac{1}{2} \log \left|x^2-2 x-5\right|+\frac{2}{\sqrt{6}} \log \left|\frac{x-1-\sqrt{6}}{x-1+\sqrt{6}}\right|\) + C

Question 6.
\(\frac{5 x+3}{\sqrt{x^2+4 x+10}}\)
Solution:
∫ \(\frac{5 x+3}{\sqrt{x^2+4 x+10}}\) dx
Let 5x + 3 = A\(\frac{d}{d x}\)(x² + 4x + 10) + B
⇒ 5x + 3 = A(2x + 4) + B
⇒ 5x + 3 = 2Ax + 4A + B
On equating the coefficients of x and constant term on both sides, we get
2A = 5 ⇒ A = \(\frac{5}{2}\)
4A + B = 3 ⇒ B = -7
⇒ 5x + 3 = \(\frac{5}{2}\)(2x + 4) – 7

Question 7.
\(\frac{1}{x \sqrt{a x-x^2}}\) [Hint : Put x = \(\frac{a}{t}\)]
Solution:
Let I = ∫\(\frac{1}{x \sqrt{a x-x^2}}\) dx
Put x = \(\frac{a}{t}\) ⇒ dx = \(-\frac{a}{t^2}\)dt

Practicing the AP Inter 2nd Year Maths Study Material and AP Inter 2nd Year Maths Chapter 7 Integrals Solutions Exercise 7c will help students to clear their doubts quickly.

Inter 2nd Year Maths Integrals Solutions Exercise 7c

Integrals Exercise 7c Solutions

I. Find the integral of the functions

Question 1.
sin²(2x + 5)
Solution:
sin²(2x + 5) = \(\frac{1-\cos 2(2 x+5)}{2}=\frac{1-\cos (4 x+10)}{2}\)
∴ ∫sin²(2x + 5) = ∫ \(\frac{1-\cos (4 x+10)}{2}\)dx
= \(\frac{1}{2} \int 1 d x-\frac{1}{2} \int \cos (4 x+10) d x\)
= \(\frac{1}{2} x-\frac{1}{2}\left(\frac{\sin (4 x+10)}{4}\right)+C\) [∵ ∫ cos x dx = sin x + C]
= \(\frac{1}{2} x-\frac{1}{8} \sin (4 x+10)+C\)

Question 2.
sin 3x cos 4x
Solution:
∫sin 3x cos 4xdx
We know that,
sin A cos B = \(\frac{1}{2}\)[(sin(A + B) + sin(A – B)]
= \(\frac{1}{2}\)∫(sin(3x + 4x) + sin(3x – 4x) dx
= \(\frac{1}{2}\)∫(sin 7x + sin(-x)) dx = \(\frac{1}{2}\)∫(sin 7x – sin x) dx
= \(\frac{1}{2}\)∫sin 7xdx – \(\frac{1}{2}\)∫sin x dx [∵ ∫ sin x dx = -cos x + C]
= \(\frac{1}{2}\) \(\left[\frac{-\cos 7 x}{7}-(-\cos x)\right]\) + C
= \(\frac{-\cos 7 x}{14}\) + \(\frac{\cos x}{2}\) + C

Question 3.
sin 4x sin 8x
Solution:
∫sin 4x sin 8x dx = \(\frac{1}{2}\) ∫ 2 sin 8x sin 4x dx
= \(\frac{1}{2}\) ∫(cos 4x – cos 12x) dx
∵ 2 Sin A Sin B = cos(A – B) – cos(A + B)
= \(\frac{1}{2}\left\{\frac{\sin 4 x}{4}-\frac{\sin 12 x}{12}\right\}\) + C [∵ ∫ cos x dx = sin x + C]

Question 4.
\(\frac{1-\cos x}{1+\cos x}\)
Solution:

Question 5.
\(\frac{\cos x}{1+\cos x}\)
Solution:

Question 6.
Find the integral of \(\frac{\sin ^3 x+\cos ^3 x}{\sin ^2 x \cos ^2 x}\)
Solution:
∫ \(\frac{\sin ^3 x+\cos ^3 x}{\sin ^2 x \cos ^2 x}\) dx
= ∫ \(\frac{\sin ^3 x}{\sin ^2 x \cos ^2 x}\) dx + ∫ \(\frac{\cos ^3 x}{\sin ^2 x \cos ^2 x}\) dx
= ∫ \(\frac{\sin x}{\sin ^2 x \cos ^2 x}\) dx + ∫ \(\frac{\cos x}{\sin ^2 x \cos ^2 x}\) dx
= ∫ tan x . sec x dx + ∫ cot x . cosec x dx
= sec x – cosec x + C
∵ ∫ sec x tan x dx = sec x + C, ∫ cosec x cot x dx = -cosec x + C

Question 7.
\(\frac{\cos 2 x+2 \sin ^2 x}{\cos ^2 x}\)
Solution:
∫ \(\frac{\cos 2 x+2 \sin ^2 x}{\cos ^2 x}\) dx = ∫ \(\frac{1-2 \sin ^2 x+2 \sin ^2 x}{\cos ^2 x}\) dx [∵ cos 2x = 1 – 2 sin²x]
= ∫ \(\frac{1}{\cos ^2 x}\) dx = ∫ sec² x dx
= tan x + C [∵ ∫ sec² x dx = tan x + C]

Question 8.
\(\frac{\cos 2 x}{(\cos x+\sin x)^2}\)
Solution:

Question 9.
sin^-1(cos x)
Solution:
\(\int \sin ^{-1}(\cos x) d x=\int \sin ^{-1} \sin \left(\frac{\pi}{2}-x\right) d x\) [∵ sin^-1 + cos^-1 t = \(\frac{\pi}{2}\) for |t| ≤ 1]
= \(\int\left(\frac{\pi}{2}-x\right) \) dx = \(\frac{\pi}{2}\) x – \(\frac{x^2}{2}\) + C [∵∫ dx = x + C; ∫xⁿ dx = \(\frac{x^{n+1}}{n+1}\) + C]

Question 10.
\(\frac{\sin ^2 x}{1+\cos x}\)
Solution:
∫ \(\frac{\sin ^2 x}{1+\cos x}\) dx
= ∫ \(\frac{\left(1-\cos ^2 x\right)}{(1+\cos x)}\) dx
= ∫ \(\frac{(1+\cos x)(1-\cos x)}{(1+\cos x)}\) dx [∵ sin²x = 1 – cos²x]
= ∫ (1 – cos x) dx = ∫ 1 dx – ∫ cos x dx
= x – sin x + C [∵∫ dx = x +C, ∫ cos x dx = sin x + C]

II. Find the integral of functions

Question 1.
cos 2x cos 4x cos 6x
Solution:
∫ cos 2x cos 4x cos 6x dx
= ∫cos 2x[\(\frac{1}{2}\)[cos(4x + 6x) + cos(4x – 6x)]] dx
[∵ 2 Cos A cos B = cos(A + B) + cos(A – B)]
= \(\frac{1}{2}\) ∫ [cos 2x cos 10x + cos 2x cos(-2x) dx
= \(\frac{1}{2}\)∫\(\frac{1}{2}\)[cos 2x cos 10x + cos² 2x] dx [∵ cos(-θ) = cos θ]
= \(\frac{1}{4}\)∫(cos 12x + cos8x + 1 + cos4x) dx
= \(\frac{1}{4}\left[\frac{\sin 12 x}{12}+\frac{\sin 8 x}{8}+x+\frac{\sin 4 x}{4}\right]\) + C [∵ ∫ cos x dx = sin x + C, ∫ sin x dx = -cosx + C]

Question 2.
sin x sin 2x sin 3x
Solution:
∫ sin x sin 2x sin 3x dx
= \(\frac{1}{2}\) ∫ (2 sin 3x sin x) sin 2x dx
[∵ Sin A sin B = cos(A – B) – cos(A + B)]
= \(\frac{1}{2}\) ∫ {cos 2x – cos 4x}sin 2x dx
= \(\frac{1}{2}\) ∫ {2 sin 2x cos 2x – 2 cos 4x sin 2x}dx
= \(\frac{1}{4}\) ∫ {(sin 4x + sin 0) – (sin 6x – sin 2x)}dx
[∵ 2 sin A cos B = sin (A + B) + sin (A – 8)
2 cos A sin B = sin (A + B) – sin (A – B)]
= \(\frac{1}{4}\) ∫ (sin 4x – sin 6x + sin 2x) dx
= \(\frac{1}{4}\left\{\frac{-\cos 4 x}{4}-\frac{(-\cos 6 x)}{6}+\frac{(-\cos 2 x)}{2}\right\}\) [∵ ∫ sin ax dx = – \(\frac{\cos a x}{a}\)]
= \(\frac{1}{4}\left\{\frac{-\cos 4 x}{4}-\frac{(-\cos 6 x)}{6}+\frac{(-\cos 2 x)}{2}\right\}\) + C

Question 3.
sin³(2x + 1)
Solution:
∫sin³(2x + 1)
= ∫sin²(2x + 1) sin(2x + 1) dx
= ∫(1 – cos²(2x + 1)) sin(2x + 1) dx [∵ sin² x = 1 – cos²]
Let cos(2x + 1) = t ⇒ -2 sin(2x + 1) dx = dt
⇒ sin(2x + 1) dx = \(\frac{-\mathrm{dt}}{2}\)
∴ ∫ sin³(2x + 1) dx = \(\frac{-1}{2} \int\left(1-t^2\right) d t=\frac{-1}{2}\left[t-\frac{t^3}{3}\right]\)
[∵ ∫ 1 dx = x + C, ∫xⁿ dx = \(\frac{x^{n+1}}{n+1}\) + C]
= \(\frac{\cos (2 x+1)}{2}+\frac{\cos ^3(2 x+1)}{6}\) + C

Question 4.
sin³x cos³ x
Solution:
∫sin³x cos³x dx
Let cos x = t ⇒ -sin x = \(\frac{\mathrm{dt}}{\mathrm{dx}}\) ⇒ dx = \(\frac{d t}{-\sin x}\)
∴ ∫ sin³ x . cos³x dx = ∫ sin³ x t³ \(\frac{d t}{-\sin x}\)
= -∫ sin³ x t³ dt = -∫ t³ (1 – cos² x)dt
= -∫ t³ (1 – t²)dt = – ∫ (t³ – t⁵) dt
= \(-\left(\frac{t^4}{4}-\frac{t^6}{6}\right)\) + C [∵ ∫xⁿ dx = \(\frac{x^{n+1}}{n+1}\) + C]
= \(\frac{\cos ^6 x}{6}-\frac{\cos ^4 x}{4}\) + C

Question 5.
sin⁴x.
Solution:
∫ sin⁴ x dx = ∫ (sin² x)² dx
= ∫ \(\left(\frac{1-\cos 2 x}{2}\right)^2\) dx [∵ sin² x = \(\frac{1-\cos 2 x}{2}\)]
= \(\frac{1}{4}\) ∫ (1 – cos 2x)² dx = \(\frac{1}{4}\) ∫ (1 + cos²2x – 2 cos 2x) dx
= \(\frac{1}{4}\) [∫ 1 dx + ∫ cos² 2x dx – 2∫ cos 2x dx]
= \(\frac{1}{4}\) [∫ 1 dx + ∫ \(\frac{(1+\cos 4 x)}{2}\) dx – 2 ∫ cos 2x dx] [cos²x = \(\frac{1+\cos 2 x}{2}\)]

Question 6.
cos⁴ x .
Solution:

Question 7.
\(\frac{\cos 2 x-\cos 2 \alpha}{\cos x-\cos \alpha}\)
Solution:
∫ \(\frac{\cos 2 x-\cos 2 \alpha}{\cos x-\cos \alpha}\) dx = ∫ \(\frac{\left(2 \cos ^2 x-1\right)-\left(2 \cos ^2 \alpha-1\right)}{(\cos x-\cos \alpha)}\) dx
= ∫ \(\frac{2 \cos ^2 x-1-2 \cos ^2 \alpha+1}{(\cos x-\cos \alpha)}\) dx [∵ cos 2x = 2 cos² – 1]
= ∫ \(\frac{2\left(\cos ^2 x-\cos ^2 \alpha\right)}{(\cos x-\cos \alpha)}\) dx = 2 ∫ \(\frac{(\cos x-\cos \alpha)(\cos x+\cos \alpha)}{(\cos x-\cos \alpha)}\) dx
= 2 [∫ cos x dx + cos α ∫ 1 dx]
= 2[sin x + cos α . x] + C = 2[sin x + x cos α] + C
[∵ cos x dx = sin x + C, ∫ dx = x + C]

Question 8.
\(\frac{\cos x-\sin x}{1+\sin 2 x}\)
Solution:

Question 9.
tan³2x sec 2x
Solution:
∫tan³2x sec 2x dx
Let sec 2x = t

Question 10.
tan⁴ x.
Solution:
Let I = ∫ tan⁴ x dx = ∫ (tan²x)² dx
I = ∫ (tan² x) (tan² x) dx
= ∫ (sec²x – 1)(tan² x) dx
= ∫ sec² x tan² x dx – ∫ tan² x dx
= ∫ sec² x tan²x dx – ∫ [sec²x – 1] dx
= ∫ sec² x tan² x dx – [∫ sec² x dx – f1 dx]dx
Now, let I₁ = ∫ sec² x tan² x dx and
I₂ = ∫ sec² x dx – ∫ 1 dx
Then, I = I₁ – I₂ …………… (1)
Put tan x ⇒ sec² x = \(\frac{d t}{d x}\) ⇒ dx = \(\frac{d t}{\sec ^2 x}\)
∴ I₁ = ∫ sec² x t² \(\frac{d t}{\sec ^2 x}\) = ∫ t² dt
= \(\frac{t^3}{3}\) + C₁ = \(\frac{\tan ^3 x}{3}\) + C₁ [∵ ∫xⁿ dx = \(\frac{x^{n+1}}{n+1}\) + C]
I₂ = ∫ sec² x dx – ∫ 1 dx = tan x – x + C₂
∴ Putting the values of I₁ and I₂ in (I), we get
I = \(\frac{\tan ^3 x}{3}\) + C₁ – (tan x – x) + C₂
⇒ I = \(\frac{\tan ^3 x}{3}\) – tan x + x + C (∵ C₁ + C₂ = C)

Question 11.
\(\frac{1}{\sin x \cos ^3 x}\)
Solution:

Question 12.
\(\frac{1}{\cos (x-a) \cos (x-b)}\)
Solution:

Question 13.
\(\frac{\sin x}{\sin (x-a)}\)
Solution:
∫ \(\frac{\sin x}{\sin (x-a)}\) dx = ∫ \(\frac{\sin \{(x-a)+a\}}{\sin (x-a)}\) dx
= ∫ \(\frac{\sin (x-a) \cos a+\cos (x-a) \sin a}{\sin (x-a)}\) dx
[∵ sin(A + B) = sin A cos B + cos A sin B]
= ∫ \(\frac{\sin (x-a) \cos a}{\sin (x-a)}\) dx + ∫ \(\frac{\cos (x-a) \sin a}{\sin (x-a)}\) dx
= cos a ∫ 1 dx + sin a ∫ cot(x – a) dx
= x cos a + sin a log|sin(x – a)| + C₁ [∵ ∫ cot x dx = log|sin x|]

Question 14.
\(\frac{\sin ^8 x-\cos ^8 x}{1-2 \sin ^2 x \cos ^2 x}\)
Solution:

Question 15.
\(\frac{1}{\cos (x+a) \cos (x+b)}\)
Solution:

Practicing the AP Inter 2nd Year Maths Study Material and AP Inter 2nd Year Maths Chapter 7 Integrals Solutions Exercise 7b will help students to clear their doubts quickly.

Inter 2nd Year Maths Integrals Solutions Exercise 7b

Integrals Exercise 7b Solutions

I. Integrate the following functions.

Question 1.
Find integral of \(\frac{2 x}{1+x^2}\)
Solution:
Let I = ∫\(\frac{2 x}{1+x^2}\) dx [∵ ∫\(\frac{1}{x}\) dx = log|x| + C]
Put 1 + x² = t
Differentiating w.r.t ‘x’, we get 2x dx = dt
∴ I = ∫\(\frac{1}{t}\) dt = log |t| + C
= log |1 + x²| + C = log(1 + x²) + C

Question 2.
Find integral of \(\frac{(\log x)^2}{x}\)
Solution:
Let I = ∫\(\frac{(\log x)^2}{x}\) dx [∵ ∫xⁿ dx = \(\frac{x^{n+1}}{n+1}\) + C]
Put log x = t
Differentiating w.r.t.x, we get \(\frac{1}{x}=\frac{d t}{d x}\) ⇒ dx = xdt
∴ I = ∫t² dt = \(\frac{t^3}{3}\) + C = \(\frac{(\log x)^3}{3}\) + C

Question 3.
Find integral of \(\frac{1}{x+x \log x}\)
Solution:
Let I = ∫\(\frac{1}{x+x \log x}\) dx [∵ ∫\(\frac{1}{x}\) dx = log|x| + C]
Put 1 + log x = t
Differentiating w.r.t.x, we get \(\frac{1}{x}=\frac{d t}{d x}\) ⇒ \(\frac{1}{x}\) dx = xdt
∴ I = ∫\(\frac{1}{t}\) dt = log |t| + C
= log |1 + log x| + C

Question 4.
Find integral of sin x sin(cos x)
Solution:
Let I = ∫sin x sin(cos x)
Put cos x = t [∵ ∫ sin x dx = -cos x + C]
On differentiating w.r.t.x, we get
-sin x = \(\frac{d t}{d x}\) ⇒ dx = \(\frac{\mathrm{dt}}{-\sin x}\)
∴ I = ∫ sin x sin(t) = \(\frac{\mathrm{dt}}{-\sin x}\) = -∫sin t dt
= -(-cos t) + C = cos(cos x) + C

Question 5.
Find integral of sin(ax + b) cos(ax + b)
Solution:
Let I = ∫sin(ax + b) cos(ax + b) dx
= \(\int \frac{\sin 2(a x+b)}{2} d x\) [∵ ∫ sin x dx = -cos x + C]
= \(\frac{1}{2}\left[\frac{-\cos 2(a x+b)}{2 a}\right]+\) + C
= \(\frac{-1}{4 a}\)cos2(ax + b) + C.

Question 6.
Find integral of \(\sqrt{a x}+b\)
Solution:
\(\sqrt{a x}+b\) dx = ∫ (ax + b)^1/2 dx
= \(\frac{(a x+b)^{\frac{1}{2}+1}}{a\left(\frac{1}{2}+1\right)}\) + C [∵ ∫ (ax + b)ⁿ dx = \(\frac{(a x+b)^{n+1}}{a(n+1)}\)]
= \(\frac{(a x+b)^{3 / 2}}{a\left(\frac{3}{2}\right)}\) + C
= \(\frac{2}{3 a}\) (ax + b)^3/2 + C

Question 7.
Find integral of \(x \sqrt{x+2}\)
Solution:

Question 8.
Find integral of \(x \sqrt{1+2 x^2}\)
Solution:
Let I = ∫\(x \sqrt{1+2 x^2}\) dx
Put 1 + 2x² = t
On differentiating w.r.t. x, we get
4x = \(\frac{d t}{d x}\) ⇒ dx = \(\frac{d t}{d x}\)
∴ I = \(\int \mathrm{x} \sqrt{\mathrm{t}} \frac{\mathrm{dt}}{4 \mathrm{x}}=\frac{1}{4} \int \sqrt{\mathrm{t}} \mathrm{dt}=\frac{1}{4} \int \mathrm{t}^{1 / 2} \mathrm{dt}\)
= \(\frac{1}{4} \frac{t^{(1 / 2)+1}}{(1 / 2)+1}\) + C = \(\frac{1}{4} \cdot \frac{2}{3} \cdot t^{3 / 2}\) + C [∵ ∫xⁿ dx = \(\frac{x^{n+1}}{n+1}\) + C]
= \(\frac{1}{6}\) (1 + 2x²)^3/2 + C

Question 9.
Find integral of (4x + 2)\(\sqrt{x^2+x}+1\)
Solution:
Put I = ∫(4x + 2)\(\sqrt{x^2+x}+1\) dx
Let x² + x + 1 = t
On differentiating w.r.t.x, we get
2x + 1 =\(\frac{d t}{d x}\) ⇒ dx = \(\frac{\mathrm{dt}}{(2 \mathrm{x}+1)}\)
∴ I = \(\int(4 \mathrm{x}+2) \sqrt{\mathrm{t}} \frac{\mathrm{dt}}{(2 \mathrm{x}+1)}\) [∵ ∫xⁿ dx = \(\frac{x^{n+1}}{n+1}\) + C]
= \(\int 2(2 x+1) \sqrt{t} \frac{d t}{(2 x+1)}\) = \(2 \int \sqrt{t} d t\)
= \(2 \frac{t^{(1 / 2)+1}}{(1 / 2)+1}\) + C
= \(\frac{4}{3}\)(x² + x + 1)^3/2 + C

Question 10.
Find integral of \(\frac{1}{x-\sqrt{x}}\)
Solution:
\(\int \frac{1}{x-\sqrt{x}} d x=\int \frac{1}{\sqrt{x}(\sqrt{x}-1)} d x\)
Put \(\sqrt{x}-1\) = t
⇒ Differentiating w.r.t. x, we get
\(\frac{1}{2 \sqrt{x}}=\frac{d t}{d x}\) ⇒ dx = \({2 \sqrt{x}}\) dt [∵ ∫\(\frac{1}{x}\) dx = log|x| + C]
∴ \(\int \frac{1}{\sqrt{x}(\sqrt{x}-1)} d x\) = \(\int \frac{1}{\sqrt{x} t} 2 \sqrt{x} d t\)
= \(\int \frac{2}{t} d t\) = 2 . log|t| + C
= \(2 \log |\sqrt{x}-1|+C\)

Question 11.
Find integral of \(\frac{x}{\sqrt{x+4}}\), x > 0
Solution:

Question 12.
Find integral of (x³ – 1)^1/3x⁵
Solution:
I = ∫(x³ – 1)^1/3x⁵ dx = ∫(x³ – 1)^1/3x³x² dx
Let x³ – 1 = t ⇒ x³ = t + 1
Differentiating w.r.t. x, we get
= \(\frac{1}{3}\)∫(x³ – 1)^1/3x³(3x² dx) …..(i) [∵ \(\frac{d}{dx}\)(x³ – 1) = 3x²]
3x² = \(\frac{\mathrm{dt}}{\mathrm{dx}}\) ⇒ dx = \(\frac{d t}{3 x^2}\)
∴ ∫ (x³ – 1)^1/3 x³ . x² dx
∫ t^1/3 (t + 1)x² \(\frac{d t}{3 x^2}\)

Question 13.
Find integral of \(\frac{x^2}{\left(2+3 x^3\right)^3}\)
Solution:

Question 14.
Find integral of \(\frac{1}{x(\log x)^m}\), x > 0, m ≠ 1
Solution:
\(\int \frac{1}{x(\log x)^m} d x\)
Let log x = t ⇒ \(\) ⇒ dx = xdt
∴ \(\int \frac{1}{x(\log x)^m} d x\) = \(\int \frac{1}{\mathrm{x}(\mathrm{t})^{\mathrm{m}}}\) x dt = ∫ t^-m dt
= \(\frac{\mathrm{t}^{-\mathrm{m}+1}}{-\mathrm{m}+1}\) + C [∵ ∫xⁿ dx = \(\frac{x^{n+1}}{n+1}\) + C]
= \(\frac{(\log x)^{1-m}}{1-m}\) + C

Question 15.
Find integral of \(\frac{x}{9-4 x^2}\)
Solution:
\(\int \frac{x}{9-4 x^2} d x\)
Let 9 – 4x² = t ⇒ -8x = \(\frac{\mathrm{dt}}{\mathrm{dx}}\) ⇒ dx = \(\frac{d t}{-8 x}\)
∴ \(\int \frac{x}{9-4 x^2} d x\) = \(\int \frac{x}{t} \frac{d t}{-8 x}\)
= \(\frac{1}{-8} \int \frac{1}{t} d t\)
= \(\frac{1}{-8}\) log|t| + C [∵ ∫\(\frac{1}{x}\) dx = log|x| + C]
= \(\frac{1}{-8}\) log|9 – 4x²| + C

Question 16.
Find integral of e^2x+3 dx
Solution:
∫e^2x+3 dx
Let 2x + 3 = t ⇒ 2 = \(\frac{\mathrm{dt}}{\mathrm{dx}}\) ⇒ dx = \(\frac{d t}{2}\)
∴ ∫e^(2x+3) dx = \(\int e^t \frac{d t}{2}\)
= \(\frac{1}{2}\)∫e^t dt
= \(\frac{1}{2}\)(e^t) + C [∵ ∫ e^x dx = e^x + C]
= \(\frac{1}{2}\)e^(2x+3) + C

Question 17.
Find integral of \(\frac{x}{e^{x^2}}\)
Solution:
\(\int \frac{x}{e^{x^2}} d x\)
Let x² = t ⇒ 2x = \(\frac{\mathrm{dt}}{\mathrm{dx}}\) ⇒ dx = \(\frac{\mathrm{dt}}{2 \mathrm{x}}\)
∴ \(\int \frac{x}{e^{x^2}} d x=\int \frac{x}{e^t} \frac{d t}{2 x}\)
= \(\frac{1}{2}\) ∫e^-t dt = \(\frac{-1}{2}\) e^-t + C [∵ ∫ e^x dx = e^x + C]
= –\(\frac{1}{2}\) e^-x2 + C

Question 18.
Find integral of \(\frac{e^{\tan -x}}{1+x^2}\)
Solution:
\(\int \frac{e^{\tan ^{-1} x}}{1+x^2} d x\)
Let tan^-1 x = t ⇒ \(\frac{1}{1+x^2}\) = \(\frac{\mathrm{dt}}{\mathrm{dx}}\) ⇒ dx = (1 + x²)dt
∴ \(\int \frac{e^{\tan ^{-1} x}}{1+x^2}\) dx = \(\int \frac{e^t}{1+x^2}\left(1+x^2\right) d t\)
= ∫e^t dt = e^t + C [∵ ∫ e^x dx = e^x + C]
= e^tan-1x + C

Question 19.
Find integral of \(\frac{e^{2 x}-1}{e^{2 x}+1}\)
Solution:
\(\int \frac{e^{2 x}-1}{e^{2 x}+1} d x\) = \(\int \frac{e^{2 x}-1}{e^{2 x}+1} d x\)
= \(\int \frac{e^x\left(e^x-\frac{1}{e^x}\right)}{e^x\left(e^x+\frac{1}{e^x}\right)} d x\) = \(\int \frac{e^x-e^{-x}}{\left(e^x+e^{-x}\right)} d x\)
Let e^x + e^-x = t ⇒ e^x – e^-x = \(\frac{\mathrm{dt}}{\mathrm{dx}}\) ⇒ dx = \(\frac{d t}{e^x-e^{-x}}\)
∴ \(\int \frac{e^{2 x}-1}{e^{2 x}+1} d x\) =\( \int \frac{e^x-e^{-x}}{t} \cdot \frac{d t}{e^x-e^{-x}}\)
= \(\int \frac{1}{\mathrm{t}} \mathrm{dt}\)
= log|t| + C [∵ ∫\(\frac{1}{x}\) dx = log|x| + C]
= log |e^x + e^-x| + C

Question 20.
Find integral of \(\frac{e^{2 x}-e^{-2 x}}{e^{2 x}+e^{-2 x}}\)
Solution:
\(\int \frac{e^{2 x}-e^{-2 x}}{e^{2 x}+e^{-2 x}} d x\)
Let e^2x + e^-2x = t [∵ ∫\(\frac{1}{x}\) dx = log|x| + C]
⇒ (2e^2x – 2e^-2x) = \(\frac{\mathrm{dt}}{\mathrm{dx}}\) ⇒ dx = \(\frac{d t}{2\left(e^{2 x}-e^{-2 x}\right)}\)
∴ \(\int \frac{e^{2 x}-e^{-2 x}}{e^{2 x}+e^{-2 x}} d x\) = \(\int \frac{e^{2 x}-e^{-2 x}}{t} \frac{d t}{2\left(e^{2 x}-e^{-2 x}\right)}\)
= \(\frac{1}{2} \int \frac{1}{t} d t\) = \(\frac{1}{2}\) log|t| + C
= \(\frac{1}{2}\) log|e^2x + e^-2x| + C

Question 21.
Find integral of tan²(2x – 3)
Solution:
∫ tan²(2x – 3) = ∫ sec²(2x – 3) dx – ∫ 1 dx [∵ tan²x = sec²x – 1]
Put 2x – 3 = t ⇒ 2 dx = dt ⇒ dx = \(\frac{1}{2}\) dt
∴ ∫ tan²(2x – 3)dx = ∫ sec² (2x – 3)dx – ∫ dx
= \(\frac{1}{2}\) ∫ sec²t dt – ∫ 1 dx
= \(\frac{1}{2}\) tan t – x + C [∵ ∫ sec²x dx = tan x + C; ∫ dx = x + C]
= \(\frac{1}{2}\) tan(2x – 3) – x + C

Question 22.
Find integral of sec²(7 – 4x)
Solution:
∫ sec²(7 – 4x) dx
Let 7 – 4x = t ⇒ -4 = \(\frac{\mathrm{dt}}{\mathrm{dx}}\) ⇒ dx = \(\frac{\mathrm{dt}}{-4}\)
∴ ∫sec²(7 – 4x)dx = ∫sec² t \(\frac{\mathrm{dt}}{-4}\)
= \(\frac{-1}{4}\)(tan t) + C [∵ ∫ sec²x dx = tan x + C]
= –\(\frac{-1}{4}\)tan(7 – 4x) + C

Question 23.
Find integral of \(\frac{\sin ^{-1} x}{\sqrt{1-x^2}}\)
Solution:

Question 24.
Find integral of \(\frac{2 \cos x-3 \sin x}{6 \cos x+4 \sin x}\)
Solution:

Question 25.
Find integral of \(\frac{1}{\cos ^2 x(1-\tan x)^2}\)
Solution:

Question 26.
Find integral of \(\frac{\cos \sqrt{x}}{\sqrt{x}}\)
Solution:
\(\int \frac{\cos \sqrt{x}}{\sqrt{x}}\) dx
Let \(\sqrt{x}\) = t ⇒ \(\frac{1}{2 \sqrt{x}}\) = \(\frac{\mathrm{dt}}{\mathrm{dx}}\) ⇒ dx = \(2 \sqrt{x}\) dt
∴ \(\int \frac{\cos \sqrt{x}}{\sqrt{x}} d x=\int \frac{\cos t}{\sqrt{x}} 2 \sqrt{x} d t\)
= 2∫cos t dt = 2 sin t + C [∵ ∫ cos x dx = sin x + C]
= 2 sin\(\sqrt{x}\) + C

Question 27.
Find integral of \(\sqrt{\sin 2 x} \cos 2 x\)
Solution:
\(\int \sqrt{\sin 2 x} \cos 2 x d x\)
Let sin 2x = t ⇒ 2 cos 2x dx = \(\frac{\mathrm{dt}}{\mathrm{dx}}\) ⇒ dx = \(\frac{\mathrm{dt}}{2 \cos 2 \mathrm{x}}\)
∴ \(\int \sqrt{\sin 2 x} \cos 2 x d x\) = \(\int \sqrt{t} \cos 2 x \frac{d t}{2 \cos 2 x}\)
= \(\frac{1}{2} \int \sqrt{t} d t\)
= \(\frac{1}{2} \frac{t^{\frac{1}{2}+1}}{\left(\frac{1}{2}+1\right)}+C\) = \(\frac{1}{3}\)t^3/2 + C [∵ ∫xⁿ dx = \(\frac{x^{n+1}}{n+1}\) + C]
= \(\frac{1}{3}\) (sin 2x)^3/2 + C

Question 28.
Find integral of \(\frac{\cos x}{\sqrt{1+\sin x}}\)
Solution:
\(\int \frac{\cos x}{\sqrt{1+\sin x}} d x\) = ∫ (1 + sin x)^-1/2 cos x dx
Let 1 + sin x = t ⇒ cos x = \(\frac{\mathrm{dt}}{\mathrm{dx}}\) ⇒ dx = \(\frac{d t}{\cos x}\)
∴ ∫ (1 + sin x)^-1/2 cos x dx = ∫ (t)^-1/2 cos x \(\frac{d t}{\cos x}\)
= \(\frac{t^{-1 / 2+1}}{\left(-\frac{1}{2}+1\right)}+C\) [∵ ∫xⁿ dx = \(\frac{x^{n+1}}{n+1}\) + C]
= 2t^1/2 + C = \(2 \sqrt{1+\sin x}+C\)

Question 29.
Find integral of cot x log sin x
Solution:
∫cot x log sin x dx
Let log sin x = t ⇒ \(\frac{1}{\sin x}\) cos x = \(\frac{\mathrm{dt}}{\mathrm{dx}}\)
⇒ cot x = \(\frac{\mathrm{dt}}{\mathrm{dx}}\) ⇒ dx = \(\frac{d t}{\cos x}\)
∴ ∫cot x log sin x dx = ∫ cot x . t \(\frac{d t}{\cos x}\)
= ∫ t dt = \(\frac{t^2}{2}\) + C [∵ ∫xⁿ dx = \(\frac{x^{n+1}}{n+1}\) + C]
= \(\frac{(\log \sin x)^2}{2}+C\)

Question 30.
Find integral of \(\frac{\sin x}{1+\cos x}\)
Solution:
Let I = ∫\(\frac{\sin x}{1+\cos x}\) dx
Put 1 + cosx = t
⇒ -sin x = \(\frac{\mathrm{dt}}{\mathrm{dx}}\) ⇒ dx = \(\frac{d t}{-\sin x}\)
∴ I = \(\int \frac{\sin x}{t} \times \frac{d t}{-\sin x}=-\int \frac{1}{t} d t\)
= – log |t | + C [∵ ∫\(\frac{1}{x}\) dx = log|x| + C]
= -log|1 + cos x| + C

Question 31.
Find integral of \(\frac{\sin x}{(1+\cos x)^2}\)
Solution:
Let I = ∫ \(\frac{\sin x}{(1+\cos x)^2}\) dx
Put 1 + cosx = t ⇒ -sin x = \(\frac{\mathrm{dt}}{\mathrm{dx}}\) ⇒ dx = \(\frac{d t}{-\sin x}\)
∴ I = \(\int \frac{\sin x}{(1+\cos x)^2} d x=\int \frac{\sin x}{t^2} \times \frac{d t}{-\sin x}=-\int \frac{1}{t^2} \cdot d t\)
= -∫ t^-2 dt [∵ ∫xⁿ dx = \(\frac{x^{n+1}}{n+1}\) + C]
= \(\frac{-t^{-2+1}}{-2+1}+C\) = \(\frac{1}{t}\) + C
= \(\frac{1}{1+\cos x}\) + C

Question 32.
Find integral of \(\frac{(1+\log x)^2}{x}\)
Solution:
∫ \(\frac{(1+\log x)^2}{x}\) dx
Let 1 + log x = t ⇒ \(\frac{1}{x}\) = \(\frac{\mathrm{dt}}{\mathrm{dx}}\) ⇒ dx = x dt
∴ \(\int \frac{(1+\log x)^2}{x} d x=\int \frac{t^2}{x} x d t=\int t^2 d t\)
= \(\frac{t^3}{3}\) + C [∵ ∫xⁿ dx = \(\frac{x^{n+1}}{n+1}\) + C]
= \(\frac{(1+\log x)^3}{3}\) + C

Question 33.
Find integral of \(\frac{(x+1)(x+\log x)^2}{x}\)
Solution:

Question 34.
Find integral of \(\frac{x^3 \sin \left(\tan ^{-1} x^4\right)}{1+x^8}\)
Solution:
Let I = ∫ \(\frac{x^3 \sin \left(\tan ^{-1} x^4\right)}{1+x^8}\) dx
Put tan^-1 x⁴ = t ⇒ \(\frac{1}{1+x^8}\) . 4x³ = \(\frac{\mathrm{dt}}{\mathrm{dx}}\) ⇒ dx = \(\frac{\left(1+x^8\right)}{4 x^3}\) dt
∴ I = \(\int \frac{x^3 \sin t}{\left(1+x^8\right)} \cdot \frac{1+x^8}{4 x^3} d t=\frac{1}{4} \int \sin t d t\)
= –\(\frac{1}{4}\) cos t + C [∵ ∫ sin x dx = -cos x + C]
= –\(\frac{1}{4}\) cos x (tan^-1 x⁴ + C)

Question 35.
Find integral of \(\frac{x^3}{\sqrt{1-x^8}}\)
Solution:
∫ \(\frac{x^3}{\sqrt{1-x^8}}\) dx = ∫ \(\frac{x^3}{\sqrt{1-\left(x^4\right)^2}}\) dx
Put x⁴ = t ⇒ 4x³dx = dt ⇒ x³ dx = \(\frac{1}{4}\) dt
∴ \(\int \frac{1 / 4 d t}{\sqrt{1-t^2}}=\frac{1}{4} \int \frac{d t}{\sqrt{1-t^2}}\)
= \(\frac{1}{4}\) sin^-1 t + C [∵ ∫ \(\frac{\mathrm{dx}}{\sqrt{1-\mathrm{x}^2}}\) = sin^-1x + C]
= \(\frac{1}{4}\) sin^-1 (x⁴) + C

Question 36.
Find integral of cos³x e^{log sin x}
Solution:
Let I = ∫ cos³x e^{log sin x} dx
cos³ xe^logsinx = cos³ x sin x
Let cos x = t ⇒ -sin x dx = dt
∴ I = -∫ t³ dt = –\(\frac{t^4}{4}\) + C [∵ ∫xⁿ dx = \(\frac{x^{n+1}}{n+1}\) + C]
= –\(\frac{\cos ^4 x}{4}\) + C

Question 37.
Find integral of e^{3log x}(x⁴ + 1)^-1
Solution:
Let I = ∫ e^{3log x}(x⁴ + 1)^-1 dx
= \(\frac{e^{\log x^3}}{\left(x^4+1\right)} d x\) = \(\int \frac{x^3}{\left(x^4+1\right)} d x\)
Put x⁴ + 1 = t ⇒ 4x³ dx = dt ⇒ dx = \(\frac{d t}{4 x^3}\)
∴ I = \(\int \frac{x^3}{t} \frac{d t}{4 x^3}=\frac{1}{4} \int \frac{1}{t} d t\)
= \(\frac{1}{4}\)log|t| + C [∵ ∫\(\frac{1}{x}\) dx = log|x| + C]
= \(\frac{1}{4}\)log|x⁴ + 1| + C

Question 38.
Find integral of f'(ax + b)[f(ax + b)]ⁿ
Solution:
Let I = ∫ f'(ax + b)[f(ax + b)]ⁿ dx
Put f(ax + b) = t ⇒ af'(ax + b) dx = dt
⇒ dx = \(\frac{\mathrm{dt}}{\mathrm{af}^{\prime}(\mathrm{ax}+\mathrm{b})}\)
∴ I = ∫ f'(ax + b)tⁿ\(\frac{\mathrm{dt}}{\mathrm{af}^{\prime}(\mathrm{ax}+\mathrm{b})}\)
= \(\frac{1}{a} \int t^n d t\)
= \(\frac{1}{a}\left(\frac{\mathrm{t}^{\mathrm{n}+1}}{\mathrm{n}+1}\right)+\mathrm{C}\) [∵ ∫xⁿ dx = \(\frac{x^{n+1}}{n+1}\) + C]
= \(\frac{1}{a} \frac{[f(a x+b)]^{n+1}}{n+1}+C\)

II.

Question 1.
Find integral of \(\frac{1}{x^2\left(x^4+1\right)^{3 / 4}}\)
Solution:

Question 2.
Find integral of \(\frac{1}{1+\cot x}\)
Solution:

Question 3.
Find integral of \(\frac{1}{1-\tan x}\)
Solution:

Question 4.
Find integral of \(\frac{\sqrt{\tan x}}{\sin x \cos x}\)
Solution:

Practicing the AP Inter 2nd Year Maths Study Material and AP Inter 2nd Year Maths Chapter 6 Application of Derivatives Solutions Exercise 6d will help students to clear their doubts quickly.

Inter 2nd Year Maths Application of Derivatives Solutions Exercise 6d

Application of Derivatives Exercise 6d Solutions

I.

Question 1.
Show that the function given by f(x) = \(\frac{\log x}{x}\), has maximum at x = e.
Solution:
The given function is f(x) = \(\frac{\log x}{x}\)
⇒ f'(x) = \(\frac{x\left[\frac{1}{x}\right]-\log x}{x^2}\) = \(\frac{1-\log x}{x^2}\)
Now, f'(x) = 0
⇒ 1 – logx = 0 ⇒ log x = 1
⇒ log x = log e
∴ x = e

∴ By second derivative test, f is the maximum at x = e.

Question 2.
The to equal sides of an isosceles triangle with fixed base h are decreasing at the rate of 3 cm per second. how fast is the area decreasing when the two equal sides are equal to the base ?
Solution:
Let ∆ABC be isosceles where BC is the base of fixed length b. Let the length of the two equal sides of ∆ABC be a.
Draw AD ⊥ BC.

Now in ∆ADC by applying the Pythagoras theorem,
we have: AD = \(\sqrt{a^2-\frac{b^2}{4}}\)
Area of triangle, A = \(\frac{1}{2} b \sqrt{a^2-\frac{b^2}{4}}\)
The rate of change of the area with respect to time (t) is given by,
\(\frac{\mathrm{dA}}{\mathrm{dt}}=\frac{1}{2} b \cdot \frac{2 a}{2 \sqrt{a^2-\frac{b^2}{4}}} \frac{d a}{d t}=\frac{a b}{\sqrt{4 a^2-b^2}} \frac{d a}{d t}\)
It is given that the two equal sides of the triangle are decreasing at the rate of 3 cm per second.
∴ \(\frac{\mathrm{da}}{\mathrm{dt}}\) = -3cm/s
⇒ \(\frac{\mathrm{dA}}{\mathrm{dt}}=\frac{-3 \mathrm{ab}}{\sqrt{4 \mathrm{a}^2-\mathrm{b}^2}}\)
When a = b, we have \(\frac{\mathrm{dA}}{\mathrm{dt}}=\frac{-3 b^2}{\sqrt{4 a^2-b^2}}=\frac{-3 b^2}{\sqrt{3 b^2}}=-\sqrt{3} b\)
Hence, if the two equal sides are equal to the base, then the area of the triangle is decreasing at the rate of \(\sqrt{3} \mathrm{~b}\) cm² / s.

Question 3.
Find the intervals in which the function f given by f(x) = \(\frac{4 \sin x-2 x-x \cos x}{2+\cos x}\) is (i) increasing (ii) decreasing
Solution:

Now, f'(x) = 0 ⇒ cos x = 0 or cos x = 4
But cos x ≠ 4
Hence,cos x = 0 ⇒ x = \(\frac{\pi}{2}\), \(\frac{3 \pi}{2}\)
Now x = \(\frac{\pi}{2}\)and x = \(\frac{3\pi}{2}\) divide(0, 2π) into three disjoint intervals i.e.,,
(0, \(\frac{\pi}{2}\)), (\(\frac{\pi}{2}\), \(\frac{3\pi}{2}\)) and (\(\frac{3\pi}{2}\), 2π)
In intervals, (0, \(\frac{\pi}{2}\)) and (\(\frac{3\pi}{2}\), 2π), f'(x) > 0
Thus, f(x) is increasing for 0 < x < \(\frac{\pi}{2}\) and \(\frac{3\pi}{2}\) < x < 2π
In the interval (\(\frac{\pi}{2}\), \(\frac{3\pi}{2}\)) , f'(x) < 0
Thus, f(x) isdecreasing for \(\frac{\pi}{2}\) < x < \(\frac{3\pi}{2}\)

Question 4.
Find the intervals in which function f given by f(x) = x³ + \(\frac{1}{x^3}\), x ≠ 0
(i) increasing
(ii) decreasing
Solution:
Given that f(x) = x³ + \(\frac{1}{x^3}\)
⇒ f'(x) = 3x² – \(\frac{3}{x^4}\) = \(\frac{3 x^6-3}{x^4}\)
f'(x) = 0 ⇒ 3x⁶ – 3 = 0
⇒ x⁶ = 1
⇒ x = ±1
Now, the points x = 1 and x = -1
Divide the real line into three disjoint intervals
i.e., ,(-∞, -1), (-1, 1) and (1, ∞)
In intervals (-∞, -1) and (1, ∞) i.e., when x < -1 and x > 1, f'(x) > 0
Thus, when x < -1 and x > 1, f is increasing.
In interval (-1, 1) i.e., -1 < x < 1, f'(x) < 0.
Thus, when -1 < x < 1, f is decreasing.

Question 5.
Find the points at which the function f given by f (x) = (x – 2)⁴ (x + 1)³ has
(i) local maxima
(ii) local minima
(iii) point of inflexion
Solution:
The given function is f (x) = (x – 2)⁴ (x + 1)³
f'(x) = 4(x – 2)³ (x + 1)³ + 3(x + 1)² (x – 2)⁴
= (x – 2)³ (x + 1)² [4(x + 1) + 3(x – 2)] = (x – 2)³ (x + 1)² (7x – 2)
Now, f'(x) = 0 ⇒ x = -1, x = \(\frac{2}{7}\), x = 2
For values of close to \(\frac{2}{7}\) and to the left of \(\frac{2}{7}\), f'(x) > 0
Also, for values of x close to \(\frac{2}{7}\) and to the right of \(\frac{2}{7}\), f'(x) < 0.
Thus, x = \(\frac{2}{7}\) is the point of local maxima.
Now, for values of x close to 2 and to the left of 2, f'(x) < 0 Also, for values of close to 2 and to the right of 2, f'(x) > 0.
Thus, x = 2 is the point of local minima.
Now, as the value of varies through -1, f'(x) does not change its sign.
Thus, x = -1 is the point of inflexion.

Question 6.
Find the absolute maximum and minimum values of the function f given by f (x) = cos² x + sin x, x ∈ [0, π]
Solution:
Given that f (x) = cos² x + sin x, x ∈ [0, π]
f'(x) = 0 ⇒ -2sin x cos x + cos x = 0
⇒ cos x = 2 sin x cos x ⇒ cos x (2 sin x – 1) = 0
⇒ sin x = \(\frac{1}{2}\) or cos x = 0
⇒ x = \(\frac{\pi}{6}\) or \(\frac{\pi}{2}\) ∵ x ∈ [0, π]
Now we evaluate the value of f at critical points x = \(\frac{\pi}{6}\), \(\frac{\pi}{2}\) and at the, end points of the interval [0, π] i.e., at x = 0 and x = π, we have.
(i) f\(\left(\frac{\pi}{6}\right)\) = c0s²\(\left(\frac{\pi}{6}\right)\) + sin \(\left(\frac{\pi}{6}\right)\) = \(\left(\frac{\sqrt{3}}{2}\right)^2+\frac{1}{2}=\frac{5}{4}\)
(ii) f(0) = cos²(0) + sin(0) = 1 + 0 = 1
(iii) f(π) = cos²(π) + sin(π) = (-1)² + 0 = 1
(iv) f\(\left(\frac{\pi}{2}\right)\) = cos²\(\left(\frac{\pi}{2}\right)\) + sin\(\left(\frac{\pi}{2}\right)\) + sin\(\left(\frac{\pi}{2}\right)\) = 0 + 1 = 1
Hence, the absolute maximum value of f is 5/4 occurring at x = π/6 and the absolute minimum value of f is 1 occurring at x = 0, π/2, π.

Question 7.
Let f be a function defined on [a, b] such that f'(x) > 0. for all x ∈ (a, b). Then prove that f is an increasing function on (a, b).
Solution:
We have to prove that function is always increasing i.e.,
f(x₂) > f(x₁) for all x₂ > x₁ [where x₁, x₂ ∈ [a, b]]
Let x₁ and x₂ be two numbers in the interval [a, b]
i.e., x₁, x₂ ∈ [a, b] and x₂ > x₁.
Consider the interval [x₁, x₂]
Function f is continuous as well as differential in [x₁, x₂] as it is continuous and differential in [a, b].
Using mean value theorem, ∃ c ∈ [x₁, x₂] such that
f'(c) = \(\frac{f\left(x_2\right)-f\left(x_1\right)}{x_1-x_2}\) ……………. (1)
Given that f'(x) > 0 ∀ x ∈ (a, b),
∴ f'(c) > 0 ∀ x ∈ [x₁, x₂]
⇒ \(\frac{f\left(x_2\right)-f\left(x_1\right)}{x_1-x_2}\) > 0
⇒ f(x₂) – f(x₁) > 0
⇒ f(x₂) >f(x₁)
⇒ f(x₁) > f(x₂)
Now, for the two points x₁, x₂ ∈ (a, b), where x₂ > x₁, we have f(x₂) > f(x₁)
Hence, the function f is increasing in the interval [a, b].

II.

Question 1.
Find the maximum area of an isosceles triangle inscribed in the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}\) = 1 with its vertex at one end of the major axis.
Solution:
The given ellipse is \(\frac{x^2}{a^2}+\frac{y^2}{b^2}\) = 1
Let the major axis be along the x-axis.
Let ABC be the triangle inscribed in the ellipse where vertex C is at (a, 0)
Let A = (-a cos θ, b sin θ) and B = (-a cos θ, -b sin θ)
so that AB = 2b sin θ
Area of ∆ABC is
f(θ) = b sin θ(a + a cos θ) = ab sin θ(1 + cos θ)
f'(θ) = ab[-sin²θ + cos θ(1 + cos θ)]
Now f'(θ) = 0 ⇒ cos θ(1 + cos θ) = sin²θ
⇒ cos θ = 1 – cos θ ⇒ cos θ = 1/2

f(θ) is maximum when θ = \(\frac{\pi}{3}\) and the maximum value is f\(\left(\frac{\pi}{3}\right)\) = ab\(\frac{\sqrt{3}}{2}\left(1+\frac{1}{2}\right)\) = \(\frac{3 \sqrt{3}}{4}\) ab

Question 2.
A tank with rectangular base and rectangular sides, open at the top is to be constructed so that its depth is 2 m and volume is 8 m³. If building of tank costs Rs 70 per sq metres for the base and Rs 45 per square metre for sides. What is the cost of least expensive tank?
Solution:
Let l, b, and h represent the length, breadth, and height of the tank respectively.
Then, we have height , h = 2m and
Volume of the tank, V = 8m³
Volume of the tank V = lbh
⇒ 8 = l × b × 2
⇒ lb = 4
⇒ b = \(\frac{4}{l}\)
Now, area of the base, lb = 4
Area of the 4 walls, A = 2h(l + b)
⇒ A = 4(l + \(\frac{4}{l}\)) ⇒ \(\frac{\mathrm{dA}}{\mathrm{dl}}=4\left(1-\frac{4}{l^2}\right)\)
Now, \(\frac{\mathrm{dA}}{\mathrm{dl}}\) = 0 ⇒ (1 – \(\frac{4}{l^2}\)) = 0
⇒ l² = 4
⇒ l = ± 2
However, the length cannot be negative,
∴ we have l = 2
Hence, b = \(\frac{4}{l}\) = \(\frac{4}{2}\) = 2
Now, \(\frac{\mathrm{d}^2 \mathrm{~A}}{\mathrm{dl}^2}=\frac{32}{l^3}\)
When, l = 2
Then, \(\frac{\mathrm{d}^2 \mathrm{~A}}{\mathrm{~dl}^2}=\frac{32}{8}\) = 4 > 0
Thus, by second derivative test, the area is the minimum when l = 2
We have l = b = h = 2
∴ Cost of building the base in ₹ is 70(lb) = 70(4) = ₹ 280
Cost of building the walls in ₹ is 2h(l + b) × 45
= 2 × 2(2 + 2) × 45 = ₹ 720
Required total cost is ₹ is 280 + 720 = ₹ 1000
Thus, the total cost of the tank will be ₹ 1000.

Question 3.
The sum of the perimeter of a circle and square is k, where k is some constant. Prove that the sum of their areas is least when the side of square is double the radius of the circle.
Solution:
Let V be the radius of the circle and ‘a’ be the side of the square .
Then, we have 2πr + 4a = k (where k is a constant)
⇒ a = \(\frac{\mathrm{k}-2 \pi \mathrm{r}}{4}\)
The sum of the areas of the circle and the square (A) is given by,
A = πr² + a² = πr² + \(\frac{(\mathrm{k}-2 \pi \mathrm{r})^2}{16}\)
Now, \(\frac{\mathrm{dA}}{\mathrm{dr}}\) = 0
⇒ 2πr – \(\frac{\pi(\mathrm{k}-2 \pi \mathrm{r})}{4}\) = 0 ⇒ 2πr = \(\frac{\pi(\mathrm{k}-2 \pi \mathrm{r})}{4}\)
⇒ 8r = k – 2πr
⇒ 2(4 + π)r = k
⇒ r = \(\frac{\mathrm{k}}{2(4+\pi)}\)
⇒ r = \(\frac{\mathrm{k}}{2(4+\pi)}\) ………… (1)
Now, \(\frac{\mathrm{d}^2 \mathrm{~A}}{\mathrm{dr}^2}\) = 2π + \(\frac{\pi^2}{2}\) > 0
When, r = \(\frac{\mathrm{k}}{2(4+\pi)}\) ⇒ \(\frac{\mathrm{d}^2 \mathrm{~A}}{\mathrm{dr}^2}\) > 0
The sum of the areas is least when, r = \(\frac{\mathrm{k}}{2(4+\pi)}\)
a = \(\frac{\mathrm{k}-2 \pi\left[\frac{\mathrm{k}}{2(4+\pi)}\right]}{4}=\frac{\mathrm{k}(4+\pi)-\pi \mathrm{k}}{4(4+\pi)}\)
= \(\frac{4 \mathrm{k}}{4(4+\pi)}=\frac{\mathrm{k}}{4+\pi}\)
= 2r [From (1)]
Hence, it has been proved that the sum of their areas is least when the side of the square is double the radius of the circle.

Question 4.
A window is in the form of a rectangle surmounted by a semicircular opening. The total perimeter of the window is 10 m. Find the dimensions of the window to admit maximum light through the whole opening.
Solution:
Let x and y be the length and breadth of the rectangular window.
Radius of the semicircular opening be x/2.
It is given that the perimeter of the window is 10m.

∴ By second derivative test, the area is the maximum when length is x = \(\frac{20}{\pi+4}\) m
Now, y =5 – \(\frac{20}{\pi+}\left(\frac{2+\pi}{4}\right)=5-\frac{5(2+\pi)}{\pi+4}=\frac{10}{\pi+4}\)
Hence, the required dimensions of the window to admit maximum light is given by length \(\frac{20}{\pi+4}\) m and breadth \(\frac{10}{\pi+4}\) m.

Question 5.
A point on the hypotenuse of a triangle is at distance a and b from the sides of the triangle. Show that the minimum length of the hypotenuse is \(\left(a^{\frac{2}{3}}+b^{\frac{2}{3}}\right)^{\frac{3}{2}}\)
Solution:
Let ∆ABC be right-angled traingle and right angle at B. Let AB = x, BC = y
Let P be a point on the hypotenuse of the triangle such that P is at a distance of a and b from the sides AB and BC respectively.
Let ∠C = θ then we have, AC = \(\sqrt{x^2+y^2}\)
Now, PC = b cosec θ and AP = a sec θ
AC = AP + PC ⇒ AC = b cosec θ + a sec θ ………… (1)
\(\frac{\mathrm{d}(\mathrm{AC})}{\mathrm{d} \theta}\) = -b cosec θ cot θ + a sec θ tan θ
∴ \(\frac{\mathrm{d}(\mathrm{AC})}{\mathrm{d} \theta}\) = 0 ⇒ a sec θ tan θ = b cosec θ cot θ

It can be clearly shown that \(\frac{\mathrm{d}^2}{\mathrm{~d} \theta^2}\)(AC) > 0
when tan θ = (b / a)^1/3
∴ By second derivative test, the length of the hypotenuse is minimum when tan θ = (b / a)^1/3
Now, if tan θ = (b / a)^1/3, we have
AC = \(\frac{b \sqrt{a^{2 / 3}+b^{2 / 3}}}{b^{1 / 3}}+\frac{a \sqrt{a^{2 / 3}+b^{2 / 3}}}{a^{1 / 3}}\)
= \(\sqrt{a^{2 / 3}+b^{2 / 3}}\) (a^2/3 + b^2/3)
= (a^2/3 + b^2/3)^3/2
Hence, the minimum length of the hypotenuse is (a^2/3 + b^2/3)^3/2.

Question 6.
Show that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius r is \(\frac{4 r}{3}\).
Solution:
A sphere of fixed radius (r) is given.
Let R and h be the radius and the height of the cone respectively.
The volume V of the cone is given by
V = \(\frac{1}{3}\) πR² h
Now, from the right ABCD ,
We have: BC = \(\sqrt{\mathrm{r}^2-\mathrm{R}^2}\)
⇒ H = r + \(\sqrt{\mathrm{r}^2-\mathrm{R}^2}\)


Now, when R² = \(\frac{8 r^2}{9}\), it can be shown that \(\frac{\mathrm{d}^2 \mathrm{~V}}{\mathrm{dR}^2}\) < 0
The volume is maximum when R² = \(\frac{8 r^2}{9}\)
When R² = \(\frac{8 r^2}{9}\),
Height of the cone is
H = r + \(\sqrt{r^2-\frac{8 r^2}{9}}\) = r + \(\sqrt{\frac{\mathrm{r}^2}{9}}\)
= r + \(\frac{r}{3}\) = \(\frac{4 \mathrm{r}}{3}\)
Hence, it can be seen that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius \(\frac{4 \mathrm{r}}{3}\).

Question 7.
Show that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius R is \(\frac{2 R}{\sqrt{3}}\). Also find the maximum volume.
Solution:
A sphere of fixed radius (R) is given.
Let r and h be the radius and the height of the cylinder respectively.

From the given figure, we have h = 2\(\sqrt{R^2-r^2}\)
The volume (V) of the cyclinder is given by,
V = πr²h = 2πr²\(\sqrt{R^2-r^2}\)

∴ The volume is maximum, when r² = \(\frac{2 R}{\sqrt{3}}\).
When r² = \(\frac{2 R}{\sqrt{3}}\), the height of the cylinder,
h = \(2 \sqrt{R^2-\frac{2 R^2}{3}}\) = \(\frac{2 \mathrm{R}}{\sqrt{3}}\)
Hence, the volume of the cylinder is maximum when the height of cylinder is \(\frac{2 \mathrm{R}}{\sqrt{3}}\).

Question 8.
Show that height of the cylinder of greatest volume which can be inscribed in a right circular cone of height h and semi vertical angle a is one-third that of the cone and the greatest volume of cylinder is \(\frac{4}{27}\) πh³ tan² α.
Solution:
The given right circular cone of fixed height h and semi-vertical angle (α) are given. Here, a cylinder of radius R and height H is inscribed in the cone.

∴ Then ∠GAO = α, OG = r, OA = h; OE = r and CE = H
We have, r = h tanα
Since ∆AOG is similar to ∆ CEG we have:

By second derivative test, the volume of the cylinder is the greatest when R = \(\frac{2 h}{3}\) tan α
H = \(\frac{1}{\tan \alpha}\left(\mathrm{~h} \tan \alpha-\frac{2 \mathrm{~h}}{3} \tan \alpha\right)\)
= \(\frac{1}{\tan \alpha}\left(\frac{\mathrm{~h} \tan \alpha}{3}\right)=\frac{\mathrm{h}}{3}\)
Thus, the height of the cylinder is one-third the height of the cone when the volume of the cylinder is the greatest. Now, the maximum volume of the cylinder can be obtained as
V = \(\pi\left(\frac{2 \mathrm{~h}}{3} \tan \alpha\right)^2 \frac{\mathrm{~h}}{3}=\pi\left(\frac{4 \mathrm{~h}^2}{9} \tan ^2 \alpha\right) \frac{\mathrm{h}}{3}\)
= \(\frac{4}{27}\) πh³ tan²α
Hence, the given result is proved.