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:
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L1 Q1
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L1 Q1.1
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L1 Q1.2

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’.
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L1 Q2
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L1 Q2.1

Question 3.
Evaluate \(\int \frac{1}{1+\sin x+\cos x} d x\). [(AP) Mar. ’20; (TS) ’15]
Solution:
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L1 Q3

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

Question 4.
Evaluate \(\int \frac{d x}{4 \cos x+3 \sin x}\). [(TS) Mar. ’18]
Solution:
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L1 Q4
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L1 Q4.1
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L1 Q4.2

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’
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L1 Q5
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L1 Q5.1

Question 6.
Evaluate \(\int \frac{1}{2-3 \cos 2 x} d x\). [May ’10, ’05]
Solution:
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L1 Q6
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L1 Q6.1

Question 7.
Evaluate \(\int \frac{d x}{5+4 \cos 2 x}\). [May ’13, Mar. ’11]
Solution:
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L1 Q7
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L1 Q7.1

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)
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L1 Q8
Comparing constant terms on both sides, we get
µ + γ = 7
⇒ 2 + γ = 7
⇒ γ = 5
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L1 Q8.1
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L1 Q8.2
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L1 Q8.3
where k is an integration constant.

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

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),
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L1 Q9
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L1 Q9.1
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L1 Q9.2

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),
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L1 Q10

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)
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L1 Q11
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,
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L1 Q11.1
Put 4 cos x + 5 sin x = t
(-4 sin x + 5 cos x) dx = dt
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L1 Q11.2

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
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L1 Q12

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

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) – \(\frac{1}{2}\)
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L1 Q13

Question 14.
Evaluate \(\int \sqrt{\frac{5-x}{x-2}} d x\). [(AP) May ’19, (TS) ’17]
Solution:
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L1 Q14
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L1 Q14.1
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L1 Q14.2
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L1 Q14.3

Question 15.
Evaluate \(\int(6 x+5) \sqrt{6-2 x^2+x} d x\). [(AP) & (TS) May ’18]
Solution:
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L1 Q15
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L1 Q15.1

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) + \(\frac{1}{2}\)
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L1 Q16

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) – \(\frac{5}{4}\)
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L1 Q17

Question 18.
Evaluate \(\int \frac{d x}{(1+x) \sqrt{3+2 x-x^2}}\). [(TS) Mar. ’20; May ’14, ’05]
Solution:
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L1 Q18
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L1 Q18.1

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

Question 19.
Evaluate \(\int \frac{d x}{(x+1) \sqrt{2 x^2+3 x+1}}\). [Mar. ’18 (TS)]
Solution:
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L1 Q19

Question 20.
Evaluate the Reduction formula for In = ∫sinnx dx and hence find ∫sin4x dx, ∫sin5x dx. [(TS) Mar. ’20, May 18; (AP) May ’19, 15; Mar. ’17; Mar. ’14, ’13]
Solution:
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L1 Q20
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L1 Q20.1

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:
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L1 Q21
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L1 Q21.1

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:
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L1 Q22
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L1 Q22.1

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:
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L1 Q23

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:
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L1 Q24
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L1 Q24.1

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

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:
\(I_n=\int \sec ^n x d x=\int \sec ^{n-2} x \cdot \sec ^2 x d x\)
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L1 Q25

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
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L2 Q1

Question 27.
Evaluate ∫x cos-1x dx. [Mar. ’09]
Solution:
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L2 Q2
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L2 Q2.1

Question 28.
Evaluate ∫x sin-1x dx. [Mar. ’04]
Solution:
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L2 Q3
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L2 Q3.1
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L2 Q3.2

Question 29.
Evaluate \(\int \frac{d x}{x^2+x+1}\)
Solution:
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L2 Q4
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L2 Q4.1

Question 30.
Evaluate \(\int \frac{d x}{\sqrt{1+x-x^2}}\)
Solution:
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L2 Q5

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

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
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L2 Q6

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
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L2 Q7
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L2 Q7.1

Question 33.
Evaluate \(\int \frac{d x}{\left(x^2+a^2\right)\left(x^2+b^2\right)}\)
Solution:
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L2 Q8
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L2 Q8.1

Question 34.
Evaluate \(\int \frac{d x}{\left(x^2+a^2\right)^2}\)
Solution:
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L2 Q9

Question 35.
Evaluate ∫eax sin(bx + c) dx. [Mar. ’19 (TS)]
Solution:
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L2 Q10
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L2 Q10.1

Question 36.
Evaluate \(\int \sqrt{\mathbf{a}^2-x^2} d x\). [Mar. ’02]
Solution:
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L2 Q11
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L2 Q11.1

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

Question 37.
Evaluate \(\int \sqrt{1+3 x-x^2} d x\). [Mar. ’11]
Solution:
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L2 Q12

Question 38.
Evaluate \(\int \frac{d x}{\sin x+\sqrt{3} \cos x}\)
Solution:
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L2 Q13
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L2 Q13.1
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L2 Q13.2
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L2 Q13.3

Question 39.
Evaluate \(\int \frac{\tan ^{-1} x}{x^2} d x\). [May ’01]
Solution:
\(\int \frac{\tan ^{-1} x}{x^2} d x=\int \tan ^{-1} x \cdot x^{-2} d x\)
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L3 Q1

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)]
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L3 Q2

Question 41.
Evaluate ∫ex log(e2x + 5ex + 6) dx.
Solution:
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L3 Q3
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L3 Q3.1
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L3 Q3.2

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

Question 42.
Evaluate \(\int \frac{1}{(1+\sqrt{x}) \sqrt{x-x^2}} d x\)
Solution:
Put x = t2, then dx = 2t dt
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L3 Q4
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L3 Q4.1

Question 43.
Evaluate \(\int \frac{1}{(x-a)(x-b)(x-c)} d x\)
Solution:
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L3 Q5
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L3 Q5.1

Question 44.
Evaluate \(\int \frac{d x}{x(x+1)(x+2)}\)
Solution:
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L3 Q6
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L3 Q6.1

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
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L3 Q7

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

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
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L3 Q8

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 = \(\frac{5}{3}\)
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L3 Q9

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
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L3 Q10

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

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 = \(\frac{17}{13}\)
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L3 Q11
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L3 Q11.1

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 = \(\frac{1}{5}\)
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L3 Q12

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 – \(\frac{1}{3}\) = \(\frac{2}{3}\)
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L3 Q13
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L3 Q13.1

Question 52.
Evaluate \(\int \tan ^{-1} \sqrt{\frac{1-x}{1+x}} d x\)
Solution:
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L3 Q14
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L3 Q14.1

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

Question 53.
Find the Reduction formula for ∫sinmx cosnx dx for a +ve integer and n ≥ 2.
Solution:
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L3 Q15
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L3 Q15.1

Question 54.
If In = ∫(log x)n dx, then show that In = x(log x)n – nIn – 1 and find ∫(log x)4 dx.
Solution:
TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type L3 Q16
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

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