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.

sin^{4}x . cos^{3}x

Solution:

Question 10.

\(\frac{1}{4 x+5 x^{-11}}\)

Solution:

Question 11.

x^{9} . sec^{2}(x^{10})

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 cos^{2}x, 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 cos^{2}x, 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

On adding, we get

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

On adding, we get

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.

cos^{8} x . cot x

Solution:

Question 11.

tan^{5}x

Solution:

Let I = ∫ tan^{5}x dx

Question 12.

cos^{7}x

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.

sin^{5}x cos^{8}x

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: