#### Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 57 Maths Textbook Solution.

$I=\frac{1}{4} \tan ^{-1} \frac{(2 \tan x+1)}{2}+c$

Given:

$\int \frac{d x}{4 \sin ^{2} x+4 \sin x \cos x+5 \cos ^{2} x}$

Hint:

To solve this statement we have to divide numerator and denominator by cos²θ.

Solution:

$I=\int \frac{1}{4 \sin ^{2} x+4 \sin x \cos x+5 \cos ^{2} x} d x$

Dividing numerator and denominator by cos²x.

$I=\int \frac{\sec ^{2} x}{4 \tan ^{2} x+4 \sin x \cos x \frac{1}{\cos ^{2} x}+\frac{5 \cos ^{2} x}{\cos ^{2} x}} d x$

$I=\int \frac{\sec ^{2} x}{4 \tan ^{2} x+4 \tan x+5} d x$

$I=\int \frac{\sec ^{2} x}{(2 \tan x+1)^{2}+4} d x$

$\left[\because 2 \tan x+1=t, d t=2 \sec ^{2} x d x, \sec ^{2} x d x=\frac{d t}{2}\right.$

$I=\int \frac{d t\left(\frac{1}{2}\right)}{t^{2}+2^{2}}=\frac{1}{4} \int \frac{d t}{t^{2}+2^{2}}$

$I=\frac{1}{2} \times \frac{1}{2} \tan ^{-1} \frac{t}{2}+c$

$I=\frac{1}{4} \tan ^{-1} \frac{(2 \tan x+1)}{2}+c$