Inter 2nd Year

Attempting AP Inter 2nd Year Maths Study Material Chapter 1 Relations and Functions MCQ helps students become familiar with objective questions.

AP Inter 2nd Year Maths Relations and Functions MCQ

Question 1.
Let R he the relation in the set {1, 2, 3, 4} given by I
R = {(1, 2), (2, 2), (1, 1), (4,4), (1, 3), (3, 3), (3, 2)). Choose the correct answer.
1) R is reflexive and symmetric but not transitive.
2) R is reflexive and transitive but not symmetric.
3) R is symmetric and transitive but not reflexive.
4)R is an equivalence relation.
Answer:
2) R is reflexive and transitive but not symmetric.

Explanation:
Given R = {(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3) (3, 2)}
Reflexive : (a, a) ∈ R v a ∈ A
i.e.,(1, 1), (2, 2), (3, 3), (4, 4) ∈ R
⇒ R is reflexive.

Symmetric : If (a, b) ∈ R then (b, a) ∈ R (1, 2) ∈ R but (2, 1) ∉ R
⇒ R is not symmetric.

Transitive : If (a, b) & (b, c) ∉ R then (a, c) ∉ R
⇒ R is transitive.
Hence option (2) is correct.

Question 2.
Let R be the relation in the set N given by R = {(a, b): a = b – 2. b > 6}.
Choose the correct answer.
1) (2, 4) ∈ R
2) (3, 8) ∈ R
3) (6, 8) ∈ R
4) (8, 7) ∈ R
Answer:
3) (6, 8) ∈ R

Explanation:
(6, 8) ∈ R, 8 > 6 & 6 = 8 – 2
Hence option (3) is correct.

Question 3.
Let f : R → R be defined as f(x) = x⁴. Choose the correct answer.
1) f is one-one onto
2) f is many-one onto
3) f is one-one but not onto
4) f is neither one-one nor onto.
Answer:
4) f is neither one-one nor onto.

Explanation:
f is not one-one because f(1) = f(-1) = 1 two different elements – 1 & 1 have same image 1. f is not onto because negative numbers in the codomain R has no pre-image.
Hence option (4) is correct.

Question 4.
Let f : R → R be defined as f (x) = 3x. Choose the correct answer.
1) f is one-one onto
2) f is many-one onto
3) f is one-one but not onto
4) f is neither one-one nor onto.
Answer:
1) f is one-one onto

Explanation:
f is one-one because f(x₁) = f(x₂) ⇒ x₁ = x₂
f is onto because for any y ∈ R ∃ x = \(\frac{y}{3}\) ∈ R
Hence option (1) is correct.

Question 5.
Let A = {1, 2, 3}. Then number of relations containing (1, 2) and (1, 3) which are reflexive and symmetric but not transitive is
1) 1
2) 2
3) 3
4) 4
Answer:
1) 1

Explanation:
R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (1, 3), (3, 1)} is reflexive and symmetric but not transitive because (2, 1) & (1, 3) ∈ R but (2, 3) ∉ R.
Thus there is only one such relation.
Hence option (1) is correct.

Question 6.
Let A = {1, 2, 3}. Then number of equivalence relations containing (1, 2) is
1) 1
2) 2
3) 3
4) 4
Answer:
2) 2

Explanation:
The number of equivalence relations containing (1,2) is 2.
R₁ = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)}
R₂ = A × A
Hence option (2) is correct.

Question 7.
Set A has 3 elements and the set B has 4 elements. Then the number of injective mappings that can be defined from A to B :
1) 144
2) 12
3) 24
4) 64
Answer:
3) 24

Explanation:
The number of injective mappings that can be defined from A to B is ^mP_n = ⁴P₃ = 4 × 3 × 2 = 24
Hence option (3) is correct.

Question 8.
Let A = {3, 5}. Then the number of reflexive relations on set A is
1) 2
2) 4
3) 0
4) 8
Answer:
2) 4

Explanation:
The number of reflexive relations on the set A is
2^{n2 – n} = 2^{22 – 2} = 4 (∵ n(A) = 2 = n)

Question 9.
Let f : R → R be the function defined by f(x) = \(\frac{2 x-1}{2}\) and g: R → R be another function defined by g(x) = x + 2, then (gof) \(\left(\frac{3}{2}\right)\) is
1) 2
2) 3
3) \(\frac{7}{2}\)
4) \(\frac{5}{2}\)
Answer:
2) 3

Explanation:
Given f(x) = \(\frac{2 x-1}{2}\), g(x) = x + 2
(gof)\(\left(\frac{3}{2}\right)\) = g(f(\(\left(\frac{3}{2}\right)\)) = g\(\left(\frac{2\left(\frac{3}{2}\right)-1}{2}\right)\)
= g(1) = 1 + 2 = 3

Question 10.
If f(x) = \(\) then f^-1(x) is .
1) \(\frac{4}{2 x+3}\)
2) \(\frac{2 x-3}{4}\)
3) 2x + 3
4) \(\frac{4 x-3}{2}\)
Answer:
4) \(\frac{4 x-3}{2}\)

Explanation:
Let f(x) = y
⇒ \(\frac{1}{2}\) = y ⇒ 2x + 3 = 4y
⇒ 2x = 4y – 3
⇒ x = \(\frac{4 y-3}{2}\) ⇒ f^-1(y) = \(\frac{4 y-3}{2}\)
∴ f^-1(x) = \(\frac{4 x-3}{2}\)

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: