TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type

Students must practice these Maths 2B Important Questions TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type to help strengthen their preparations for exams.

TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type

Question 1.
Evaluate \(\int \frac{d x}{4+5 \sin x}\). [Mar. ’05]
Solution:

Question 2.
Evaluate \(\int \frac{d x}{5+4 \cos x}\). [Mar. ’12, ’11, ’10]
Solution:
Write tan \(\frac{x}{2}\) = t
Differentiating on both sides with respect to ‘x’.

Question 3.
Evaluate \(\int \frac{1}{1+\sin x+\cos x} d x\). [(AP) Mar. ’20; (TS) ’15]
Solution:

Question 4.
Evaluate \(\int \frac{d x}{4 \cos x+3 \sin x}\). [(TS) Mar. ’18]
Solution:

Question 5.
Find \(\int \frac{d x}{3 \cos x+4 \sin x+6}\). [May ’15 (AP); Mar. ’13]
Solution:
Let tan \(\frac{x}{2}\) = t
Differentiating on both sides with respect to ‘x’

Question 6.
Evaluate \(\int \frac{1}{2-3 \cos 2 x} d x\). [May ’10, ’05]
Solution:

Question 7.
Evaluate \(\int \frac{d x}{5+4 \cos 2 x}\). [May ’13, Mar. ’11]
Solution:

Question 8.
Evaluate \(\int \frac{\cos x+3 \sin x+7}{\cos x+\sin x+1} d x\). [(AP) Mar. ’19; May ’06]
Solution:
Let cos x + 3 sin x + 7 = \(\frac{d}{dx}\) (cos x + sin x + 1) + µ (cos x + sin x + 1) + γ
= λ(-sin x + cos x) + µ (cos x + sin x + 1) + γ …….(1)
= -λ sin x + λ cos x + µ cos x + µ sin x + µ + γ
Comparing the coefficients of cos x on both sides, we get
λ + µ = 1 ……..(2)
Comparing the coefficients of sinx on both sides, we get
-λ + µ = 3 ………(3)
Solving (2) and (3)

Comparing constant terms on both sides, we get
µ + γ = 7
⇒ 2 + γ = 7
⇒ γ = 5



where k is an integration constant.

Question 9.
Evaluate \(\int \frac{2 \sin x+3 \cos x+4}{3 \sin x+4 \cos x+5} d x\). [Mar. ’16 (AP & TS); Mar. ’14, ’11]
Solution:
2 sin x + 3 cos x + 4 = λ(3 sin x + 4 cos x + 5) + µ(3 sin x + 4 cos x + 5) + γ
= λ(3 cos x – 4 sin x) + µ(3 sin x + 4 cos x + 5) + γ …….(1)
Comparing,
Coeff. of cos x, 3 = 3λ + 4µ
⇒ 3λ + 4µ – 3 = 0 ……….(2)
Coeff. of sin x, 2 = -4λ + 3µ
⇒ 4λ – 3µ + 2 = 0 ……..(3)
Solving (2) and (3),

Question 10.
Evaluate \(\int \frac{9 \cos x-\sin x}{4 \sin x+5 \cos x} d x\). [Mar. ’17 (TS), Mar. ’08]
Solution:
9 cos x – sin x = λ(4 sin x + 5 cos x) + µ(4 sin x + 5 cos x) + γ
= λ(4 cos x – 5 sin x) + µ(4 sin x + 5 cos x) + γ ………(1)
Comparing,
Coeff. of cos x, 9 = 4λ + 5µ
⇒ 4λ + 5µ – 9 = 0 ……….(2)
Coeff. of sin x, -1 = -5λ + 4µ
⇒ -5λ + 4µ + 1 = 0 ………(3)
Coeff. of constant, 0 = γ
Solving (2) & (3),

Question 11.
Evaluate \(\int \frac{2 \cos x+3 \sin x}{4 \cos x+5 \sin x} d x\). [(TS) May ’19, ’16; (AP) Mar. ’18, ’15]
Solution:
Let 2 cos x + 3 sin x = λ \(\frac{d}{dx}\) (4 cos x + 5 sin x) + µ(4 cos x + 5 sin x) + γ
= λ(-4 sin x + 5 cos x) + µ(4 cos x + 5 sin x) + γ ………(1)
= -4λ sin x + 5λ cos x + 4µ cos x + 5µ sin x + γ
Comparing the coefficients of cos x on both sides, we get
5λ + 4µ = 2 ……..(2)
Comparing the coefficients of sinx on both sides, we get
-4λ + 5µ = 3 ……….(3)
Solving (2) and (3)

Comparing constant terms on both sides, we get X = 0
From (1),
2 cos x + 3 sin x = \(\frac{-2}{41}\) (-4 sin x + 5 cos x) + \(\frac{23}{41}\) (4 cos x + 5 sin x) + 0
= \(\frac{-2}{41}\) (-4 sin x + 5 cos x) + \(\frac{23}{41}\) (4 cos x + 5 sin x)
Now,

Put 4 cos x + 5 sin x = t
(-4 sin x + 5 cos x) dx = dt

Question 12.
Evaluate \(\int \frac{2 x+5}{\sqrt{x^2-2 x+10}} d x\). [(AP) May ’17; Mar. ’15 (TS)]
Solution:
Write 2x + 5 = A . \(\frac{d}{dx}\) (x2 – 2x + 10) + B
= A(2x – 2) + B ………(1)
Coeff. of x, 2 = 2A then A = 1
Constant, 5 = -2A + B then B = 5 + 2 = 7
From (1), 2x + 5 = (2x – 2) + 7

Question 13.
Evaluate \(\int \frac{x+1}{x^2+3 x+12} d x\). [(AP) Mar. ’17; May ’16]
Solution:
Write x + 1 = A \(\frac{d}{dx}\) [x2 + 3x + 12) + B
= A(2x + 3) + B ……..(1)
Coefficient of x, 1 = A(2) then A = \(\frac{1}{2}\)
Constant, 1 = 3A + B then B = 1 – \(\frac{3}{2}\) = \(\frac{-1}{2}\)
From (1), x + 1 = \(\frac{1}{2}\)(2x + 3) – \(\frac{1}{2}\)

Question 14.
Evaluate \(\int \sqrt{\frac{5-x}{x-2}} d x\). [(AP) May ’19, (TS) ’17]
Solution:

Question 15.
Evaluate \(\int(6 x+5) \sqrt{6-2 x^2+x} d x\). [(AP) & (TS) May ’18]
Solution:

Question 16.
Evaluate \(\int x \sqrt{1+x-x^2} d x\). [May ’12]
Solution:
Take x = A \(\frac{d}{dx}\) (1 + x – x²) + B
x = A(1 – 2x) + B …….(1)
= A – 2Ax + B
Comparing the coefficient of x on both sides, we get
-2A = 1
⇒ A = \(-\frac{1}{2}\)
Comparing constant terms on both sides, we get
A + B = 0
⇒ \(-\frac{1}{2}\) + B = 0
⇒ B = \(\frac{1}{2}\)
From (1), x = \(-\frac{1}{2}\)(1 – 2x) + \(\frac{1}{2}\)

Question 17.
Evaluate \(\int(3 x-2) \sqrt{2 x^2-x+1} d x\). [(TS) May ’15; May ’03]
Solution:
Take 3x – 2 = A \(\frac{d}{dx}\)(2x² – x + 1) + B
3x – 2 = A(4x – 1) + B …….(1)
3x – 2 = 4Ax – A + B
Comparing the coefficients of x on both sides, we get
4A = 3
⇒ A = \(\frac{3}{4}\)
Comparing the constant terms on both sides
-A + B = -2
⇒ \(-\frac{3}{4}\) + B = -2
⇒ B = \(\frac{-5}{4}\)
From (1), 3x – 2 = \(\frac{3}{4}\) (4x – 1) – \(\frac{5}{4}\)

Question 18.
Evaluate \(\int \frac{d x}{(1+x) \sqrt{3+2 x-x^2}}\). [(TS) Mar. ’20; May ’14, ’05]
Solution:

Question 19.
Evaluate \(\int \frac{d x}{(x+1) \sqrt{2 x^2+3 x+1}}\). [Mar. ’18 (TS)]
Solution:

Question 20.
Evaluate the Reduction formula for I_n = ∫sinⁿx dx and hence find ∫sin⁴x dx, ∫sin⁵x dx. [(TS) Mar. ’20, May 18; (AP) May ’19, 15; Mar. ’17; Mar. ’14, ’13]
Solution:

Question 21.
Obtain the Reduction formula for ∫cosⁿx dx for n ≥ 2 and deduce the value of ∫cos⁵x dx. [(AP) Mar. ’20, May ’18; (TS) May ’19, ’16; Mar. ’17]
Solution:

Question 22.
Find the Reduction formula of ∫tanⁿx dx for an integer n ≥ 2. And deduce the value of ∫tan⁶x dx. [(AP) Mar. ’18, ’15; May ’16; (TS) ’17; Mar. ’12; May ’13]
Solution:

Question 23.
Find the Reduction formula of ∫cotⁿx dx for an integer n ≥ 2. And deduce the value of ∫cot⁴x dx. [Mar. ’19 (TS); Mar. ’16 (AP); May ’11]
Solution:

Question 24.
Find the Reduction formula for ∫cosecⁿx dx for an integer n ≥ 2 and deduce the value of ∫cosec⁵x dx. [Mar. ’19 (AP); Mar. ’16 (TS); May ’14]
Solution:

Question 25.
Find the Reduction formula for ∫secⁿx dx for an integer n ≥ 2 and deduce the value of ∫sec⁵x dx. [(AP) May ’17; (TS) May ’15; Mar. ’04]
Solution:
\(I_n=\int \sec ^n x d x=\int \sec ^{n-2} x \cdot \sec ^2 x d x\)

Question 26.
Evaluate \(\int \frac{\sin 2 x}{a \cos ^2 x+b \sin ^2 x} d x\)
Solution:
Put a cos²x + b sin²x = t
then [a . 2 cos x (-sin x) + b . 2 sin x . cos x] dx = dt
⇒ [-a sin 2x + b sin 2x] dx = dt
⇒ (b – a) sin 2x dx = dt
⇒ sin 2x dx = \(\frac{1}{b-a}\) dt

Question 27.
Evaluate ∫x cos^-1x dx. [Mar. ’09]
Solution:

Question 28.
Evaluate ∫x sin^-1x dx. [Mar. ’04]
Solution:

Question 29.
Evaluate \(\int \frac{d x}{x^2+x+1}\)
Solution:

Question 30.
Evaluate \(\int \frac{d x}{\sqrt{1+x-x^2}}\)
Solution:

Question 31.
Evaluate \(\int \frac{\cos x}{\sin ^2 x+4 \sin x+5} d x\). [Mar. ’07, ’03]
Solution:
Put sin x = t then cos dx = dt

Question 32.
Evaluate \(\int \frac{\sin x \cos x}{\cos ^2 x+3 \cos x+2} d x\)
Solution:
Put cos x = t
⇒ -sin x dx = dt
⇒ sin x dx = -dt

Question 33.
Evaluate \(\int \frac{d x}{\left(x^2+a^2\right)\left(x^2+b^2\right)}\)
Solution:

Question 34.
Evaluate \(\int \frac{d x}{\left(x^2+a^2\right)^2}\)
Solution:

Question 35.
Evaluate ∫e^ax sin(bx + c) dx. [Mar. ’19 (TS)]
Solution:

Question 36.
Evaluate \(\int \sqrt{\mathbf{a}^2-x^2} d x\). [Mar. ’02]
Solution:

Question 37.
Evaluate \(\int \sqrt{1+3 x-x^2} d x\). [Mar. ’11]
Solution:

Question 38.
Evaluate \(\int \frac{d x}{\sin x+\sqrt{3} \cos x}\)
Solution:

Question 39.
Evaluate \(\int \frac{\tan ^{-1} x}{x^2} d x\). [May ’01]
Solution:
\(\int \frac{\tan ^{-1} x}{x^2} d x=\int \tan ^{-1} x \cdot x^{-2} d x\)

Question 40.
Evaluate \(\int \frac{1}{a \sin x+b \cos x} d x\). [May ’03]
Solution:
Let a = r cos θ, b = r sin θ then r = \(\sqrt{a^2+b^2}\)
Now a sin x + b cos x = r cos θ sin x + r sin θ cos x = r[sin (x + θ)]

Question 41.
Evaluate ∫e^x log(e^2x + 5e^x + 6) dx.
Solution:

Question 42.
Evaluate \(\int \frac{1}{(1+\sqrt{x}) \sqrt{x-x^2}} d x\)
Solution:
Put x = t², then dx = 2t dt

Question 43.
Evaluate \(\int \frac{1}{(x-a)(x-b)(x-c)} d x\)
Solution:

Question 44.
Evaluate \(\int \frac{d x}{x(x+1)(x+2)}\)
Solution:

Question 45.
Evaluate \(\int \frac{7 x-4}{(x-1)^2(x+2)} d x\)
Solution:
Let \(\frac{7 x-4}{(x-1)^2(x+2)}=\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C}{x+2}\)
⇒ \(\frac{7 x-4}{(x-1)^2(x+2)}=\frac{A(x-1)(x+2)+B(x+2)+C(x-1)^2}{(x-1)^2(x+2)}\)
⇒ 7x – 4 = A(x – 1) (x + 2) + B(x + 2) + C(x – 1)²
If x = 1,
1 = 7(1) – 4 = B(1 + 2)
⇒ 3 = 3B
⇒ B = 1
If x = -2,
7(-2) – 4 = C(-2 – 1)²
⇒ -14 – 4 = C(-3)²
⇒ -18 = 9C
⇒ C = -2
(1) ⇒ 7x – 4 = Ax² + Ax – 2A + Bx + 2B + Cx² – 2Cx + C
Comparing the coefficients of x² on both sides, we get
A + C = 0
⇒ A – 2 = 0
⇒ A = 2

Question 46.
Evaluate \(\int \frac{x^2}{(x+1)(x+2)^2} d x\)
Solution:
Let \(\frac{x^2}{(x+1)(x+2)^2}=\frac{A}{x+1}+\frac{B}{x+2}+\frac{C}{(x+2)^2}\)
⇒ \(\frac{x^2}{(x+1)(x+2)^2}=\frac{A(x+2)^2+B(x+1)(x+2)+C(x+1)}{(x+1)(x+2)^2}\)
⇒ x² = A(x + 2)² + B(x + 1)(x + 2) + C(x + 1) …..(1)
If x = -1, then (-1)² = A(-1 + 2)²
⇒ 1 = A(1)²
⇒ A = 1
If x = -2, then (-2) = C(-2 + 1)
⇒ 4 = C(-1)
⇒ 4 = -C
⇒ C = -4
(1) ⇒ x² = Ax² + 4Ax + 4A + Bx² + 3Bx + 2B + Cx + C
Comparing the coefficients of x² on both sides, we get
1 = A + B
⇒ 1 + B = 1
⇒ B = 0

Question 47.
Evaluate \(\int \frac{2 x+3}{x^3+x^2-2 x} d x\)
Solution:
Given \(\int \frac{2 x+3}{x^3+x^2-2 x} d x=\int \frac{2 x+3}{x\left(x^2+x-2\right)} d x=\int \frac{2 x+3}{x(x+2)(x-1)} d x\)
Let \(\frac{2 x+3}{x(x+2)(x-1)}=\frac{A}{x}+\frac{B}{x+2}+\frac{C}{x-1}\)
⇒ \(\frac{2 x+3}{x(x+2)(x-1)}=\frac{A(x+2)(x-1)+B x(x-1)+C x(x+2)}{x(x+2)(x-1)}\)
⇒ 2x + 3 = A(x + 2)(x – 1) + B(x – 1) x + Cx(x + 2)
If x = 0, then 2(0) + 3 = A(0 + 2) (0 – 1)
⇒ 3 = A(2)(-1)
⇒ 3 = -2A
⇒ A = \(\frac{-3}{2}\)
If x = -2, then 2(-2) + 3 = B(-2 – 1) (-2)
⇒ -4 + 3 = B(-2)(-3)
⇒ -1 = 6B
⇒ B = \(\frac{-1}{6}\)
If x = 1, then 2(1) + 3 = C(1)(1 + 2)
⇒ 5 = C(1)(3)
⇒ 3C = 5
⇒ C = \(\frac{5}{3}\)

Question 48.
Evaluate \(\int \frac{x+3}{(x-1)\left(x^2+1\right)} d x\). [May ’07]
Solution:
Let \(\frac{x+3}{(x-1)\left(x^2+1\right)}=\frac{A}{x-1}+\frac{B x+C}{x^2+1}\)
\(\frac{x+3}{(x-1)\left(x^2+1\right)}=\frac{A\left(x^2+1\right)+(B x+C)(x-1)}{(x-1)\left(x^2+1\right)}\)
x + 3 = A(x² + 1) + (Bx + C) (x – 1) ………(1)
If x = 1 then 1 + 3 = A(1² + 1)
⇒ 4 = A(2)
⇒ A = 2
from (1), x + 3 = Ax² + A + Bx² – Bx + Cx – C
Comparing x² coefficients on both sides, we get
A + B = 0
⇒ 2 + B = 0
⇒ B = -2
Comparing coefficients of x on both sides, we get
-B + C = 1
⇒ -(-2) + C = 1
⇒ 2 + C = 1
⇒ C = 1 – 2
⇒ C = -1

Question 49.
Evaluate \(\int \frac{2 x+3}{(x+3)\left(x^2+4\right)} d x\). [May ’02]
Solution:
Let \(\frac{2 x+3}{(x+3)\left(x^3+4\right)}=\frac{A}{x+3}+\frac{B x+C}{x^2+4}\)
\(\frac{2 x+3}{(x+2)\left(x^2+4\right)}=\frac{A\left(x^2+4\right)+(B x+C)(x+3)}{(x+3)\left(x^2+4\right)}\)
2x + 3 = A(x² + 4) + (Bx + C)(x + 3) …….(1)
If x = -3 then
2(-3) + 3 = A[(-3)² + 4]
⇒ -6 + 3 = A(9 + 4)
⇒ 13A = -3
⇒ A = \(\frac{-3}{13}\)
From (1),
2x + 3 = Ax² + 4A + Bx² + 3Bx + Cx + 3C
Comparing x² coefficients on both sides, we get
A + B = 0
⇒ \(\frac{-3}{13}\) + B = 0
⇒ B = \(\frac{3}{13}\)
Comparing x coefficients on both sides, we get
3B + C = 2
⇒ 3(\(\frac{3}{13}\)) + C = 2
⇒ C = 2 – \(\frac{3}{13}\)
⇒ C = \(\frac{17}{13}\)

Question 50.
Evaluate \(\int \frac{1}{(1-x)\left(4+x^2\right)} d x\)
Solution:
\(\frac{1}{(1-x)\left(4+x^2\right)}=\frac{A}{1-x}+\frac{B x+C}{x^2+4}\)
1 = A(x² + 4) + (Bx + C)(1 – x)
Put x = 1 then A = \(\frac{1}{5}\)
Coeff. of x², 0 = A – B ⇒ B = \(\frac{1}{5}\)
Constant, 1 = C + 4A ⇒ C = \(\frac{1}{5}\)

Question 51.
Evaluate \(\int \frac{d x}{x^3+1}\). [May ’03]
Solution:
\(\frac{1}{x^3+1}=\frac{1}{(x+1)\left(x^2-x+1\right)}=\frac{A}{x+1}+\frac{B x+C}{x^2-x+1}\) ……(1)
1 = A(x² – x + 1) + (Bx + C) (x + 1)
Put x = -1, 1 = A(3) ⇒ A = \(\frac{1}{3}\)
Coeff. of x², 0 = A + B ⇒ B = \(\frac{-1}{3}\)
Constant, 1 = A + C ⇒ C = 1 – \(\frac{1}{3}\) = \(\frac{2}{3}\)

Question 52.
Evaluate \(\int \tan ^{-1} \sqrt{\frac{1-x}{1+x}} d x\)
Solution:

Question 53.
Find the Reduction formula for ∫sin^mx cosⁿx dx for a +ve integer and n ≥ 2.
Solution:

Question 54.
If I_n = ∫(log x)ⁿ dx, then show that I_n = x(log x)ⁿ – nI_n – 1 and find ∫(log x)⁴ dx.
Solution:

Now I₄ = ∫(log x)⁴ dx
= x(log x)⁴ – 4I₃
= x(log x)⁴ – 4[x(log x)³ – 4I₂]
= x(log x)⁴ – 4x(log x)³ + 16I₂
= x(log x)⁴ – 4x(log x)³ + 16[x(log x)² – 2I₁]
= x(log x)⁴ – 4x(log x)³ + 16x(log x)² – 32I₁
= x(log x)⁴ – 4x(log x)³ + 16x(log x)² – 32[x(log x) – x] + c
= x(log x)⁴ – 4x(log x)³ + 16x(log x)² – 32x log x + 32x + c