# Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(A)

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 3 Indefinite Integration Ex 3.2(A) Questions and Answers.

## Maharashtra State Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(A)

I. Integrate the following functions w.r.t. x:

Question 1.
$$\frac{(\log x)^{n}}{x}$$
Solution:

Question 2.
$$\frac{\left(\sin ^{-1} x\right)^{\frac{3}{2}}}{\sqrt{1-x^{2}}}$$
Solution:
Let I = $$\int \frac{\left(\sin ^{-1} x\right)^{\frac{3}{2}}}{\sqrt{1-x^{2}}} d x$$

Question 3.
$$\frac{1+x}{x \cdot \sin (x+\log x)}$$
Solution:

Question 4.
$$\frac{x \cdot \sec ^{2}\left(x^{2}\right)}{\sqrt{\tan ^{3}\left(x^{2}\right)}}$$
Solution:

Question 5.
$$\frac{e^{3 x}}{e^{3 x}+1}$$
Solution:

Question 6.
$$\frac{\left(x^{2}+2\right)}{\left(x^{2}+1\right)} \cdot a^{x+\tan ^{-1} x}$$
Solution:

Question 7.
$$\frac{e^{x} \cdot \log \left(\sin e^{x}\right)}{\tan \left(e^{x}\right)}$$
Solution:

Question 8.
$$\frac{e^{2 x}+1}{e^{2 x}-1}$$
Solution:

Question 9.
sin4x . cos3x
Solution:

Question 10.
$$\frac{1}{4 x+5 x^{-11}}$$
Solution:

Question 11.
x9 . sec2(x10)
Solution:

Question 12.
$$e^{3 \log x} \cdot\left(x^{4}+1\right)^{-1}$$
Solution:

Question 13.
$$\frac{\sqrt{\tan x}}{\sin x \cdot \cos x}$$
Solution:
Let I = $$\int \frac{\sqrt{\tan x}}{\sin x \cdot \cos x} d x$$
Dividing numerator and denominator by cos2x, we get

Question 14.
$$\frac{(x-1)^{2}}{\left(x^{2}+1\right)^{2}}$$
Solution:

Question 15.
$$\frac{2 \sin x \cos x}{3 \cos ^{2} x+4 \sin ^{2} x}$$
Solution:

Question 16.
$$\frac{1}{\sqrt{x}+\sqrt{x^{3}}}$$
Solution:

Question 17.
$$\frac{10 x^{9}+10^{x} \cdot \log 10}{10^{x}+x^{10}}$$
Solution:

Question 18.
$$\frac{x^{n-1}}{\sqrt{1+4 x^{n}}}$$
Solution:

Question 19.
(2x + 1) $$\sqrt{x+2}$$
Solution:

Question 20.
$$x^{5} \sqrt{a^{2}+x^{2}}$$
Solution:

Question 21.
$$(5-3 x)(2-3 x)^{-\frac{1}{2}}$$
Solution:

Question 22.
$$\frac{7+4 x+5 x^{2}}{(2 x+3)^{\frac{3}{2}}}$$
Solution:

Question 23.
$$\frac{x^{2}}{\sqrt{9-x^{6}}}$$
Solution:

Question 24.
$$\frac{1}{x\left(x^{3}-1\right)}$$
Solution:

Question 25.
$$\frac{1}{x \cdot \log x \cdot \log (\log x)}$$
Solution:

II. Integrate the following functions w.r.t x:

Question 1.
$$\frac{\cos 3 x-\cos 4 x}{\sin 3 x+\sin 4 x}$$
Solution:

Question 2.
$$\frac{\cos x}{\sin (x-a)}$$
Solution:

Question 3.
$$\frac{\sin (x-a)}{\cos (x+b)}$$
Solution:

Question 4.
$$\frac{1}{\sin x \cdot \cos x+2 \cos ^{2} x}$$
Solution:
Let I = $$\int \frac{1}{\sin x \cdot \cos x+2 \cos ^{2} x} d x$$
Dividing numerator and denominator of cos2x, we get

Question 5.
$$\frac{\sin x+2 \cos x}{3 \sin x+4 \cos x}$$
Solution:
Let I = $$\int \frac{\sin x+2 \cos x}{3 \sin x+4 \cos x} d x$$
Put, Numerator = A (Denominator) + B [$$\frac{d}{d x}$$ (Denominator)]
∴ sin x+ 2 cos x = A(3 sin x + 4 cos x) + B [$$\frac{d}{d x}$$ (3 sin x + 4 cos x)]
= A(3 sin x + 4 cos x) + B (3 cos x – 4 sin x)
∴ sin x + 2 cos x = (3A – 4B) sin x + (4A + 3B) cos x
Equating the coefficients of sin x and cos x on both the sides, we get
3A – 4B = 1 …… (1)
and 4A + 3B = 2 …… (2)
Multiplying equation (1) by 3 and equation (2) by 4, we get
9A – 12B = 3
16A + 12B = 8

Question 6.
$$\frac{1}{2+3 \tan x}$$
Solution:
Let I = $$\int \frac{1}{2+3 \tan x} d x$$

Numerator = A (Denominator) + B [$$\frac{d}{d x}$$ (Denominator)]
∴ cos x = A(2 cos x + 3 sin x) + B [$$\frac{d}{d x}$$ (2 cos x + 3 sin x)]
= A (2 cos x + 3 sin x) + B (-2 sin x + 3 cos x)
∴ cos x = (2A + 3B) cos x + (3A – 2B) sin x
Equating the coefficients of cosx and sinx on both the sides, we get
2A + 3B = 1 …… (1)
and 3A – 2B = 0 ……. (2)
Multiplying equation (1) by 2 and equation (2) by 3, we get
4A + 6B = 2
9A – 6B = 0

Question 7.
$$\frac{4 e^{x}-25}{2 e^{x}-5}$$
Solution:
Let I = $$\int \frac{4 e^{x}-25}{2 e^{x}-5} d x$$
Put, Numerator = A (Denominator) + B [$$\frac{d}{d x}$$ (Denominator)]
∴ 4ex – 25 = A(2ex – 5) + B[$$\frac{d}{d x}$$ (2ex – 5)]
= A(2ex – 5) + B(2ex – 0)
∴ 4ex – 25 = (2A + 2B) ex – 5A
Equating the coefficient of ex and constant on both sides, we get
2A + 2B = 4 …….(1)
and 5A = 25
∴ A = 5
from (1), 2(5) + 2B = 4
∴ 2B = -6
∴ B = -3

Question 8.
$$\frac{20+12 e^{x}}{3 e^{x}+4}$$
Solution:

Question 9.
$$\frac{3 e^{2 x}+5}{4 e^{2 x}-5}$$
Solution:
Let I = $$\int \frac{3 e^{2 x}+5}{4 e^{2 x}-5} d x$$
Put, Numerator = A (Denominator) + B [$$\frac{d}{d x}$$ (Denominator)]
∴ 3e2x + 5 = A(4e2x – 5) + B [$$\frac{d}{d x}$$ (4e2x – 5)]
= A(4e2x – 5) + B(4 . e2x × 2 – 0)
∴ 3e2x + 5 = (4A + 8B) e2x – 5A
Equating the coefficient of e2x and constant on both sides, we get
4A + 8B = 3 …….. (1)
and -5A = 5
∴ A = -1
∴ from (1), 4(-1) + 8B = 3
∴ 8B = 7
∴ B = $$\frac{7}{8}$$

Question 10.
cos8 x . cot x
Solution:

Question 11.
tan5x
Solution:
Let I = ∫ tan5x dx

Question 12.
cos7x
Solution:

Question 13.
tan 3x tan 2x tan x
Solution:
Let I = ∫ tan 3x tan 2x tan x dx
Consider tan 3x = tan (2x + x) = $$\frac{\tan 2 x+\tan x}{1-\tan 2 x \tan x}$$
tan 3x (1 – tan 2x tan x) = tan 2x + tan x
tan 3x – tan 3x tan 2x tan x = tan 2x + tan x
tan 3x – tan 2x – tan x = tan 3x tan 2x tan x
I = ∫(tan 3x – tan 2x – tan x) dx
= ∫tan3x dx – ∫tan 2x dx – ∫tan x dx
= $$\frac{1}{3}$$ log | sec 3x| – $$\frac{1}{2}$$ log |sec 2x| – log |sec x| + c.

Question 14.
sin5x cos8x
Solution:

Question 15.
$$3^{\cos ^{2} x \cdot} \sin 2 x$$
Solution:

Question 16.
$$\frac{\sin 6 x}{\sin 10 x \sin 4 x}$$
Solution:

Question 17.
$$\frac{\sin x \cos ^{3} x}{1+\cos ^{2} x}$$
Solution: