Inter 2nd Year Maths Exercise 7j Solutions

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.