#### Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 63 Maths Textbook Solution.

$\frac{1}{\sqrt{3}} \ln \left|\frac{\sin x-\left(\frac{\sqrt{3}}{2}\right)}{\sin x+\left(\frac{\sqrt{3}}{2}\right)}\right|+c$

Hint:

To solve this given statement we will write cos²x in terms of sin²x

Given:

$\int \frac{\cos x}{\frac{1}{4}-\cos ^{2} x} d x$

Solution:

$\int \frac{\cos x}{\frac{1}{4}-\cos ^{2} x} d x$

$I=\int \frac{\cos x}{\frac{1}{4}-\left(1-\sin ^{2} x\right)} d x$

$I=\int \frac{\cos x}{-\frac{3}{4}+\sin ^{2} x} d x$

$\sin x=t \quad d t=\cos x d x$

$\therefore I=\int \frac{d t}{t^{2}-\frac{3}{4}}$

$I=\int \frac{d t}{t^{2}-\left(\frac{\sqrt{3}}{2}\right)^{2}}$

$I=\frac{1}{2\left(\frac{\sqrt{3}}{2}\right)} \ln \left|\frac{t-\left(\frac{\sqrt{3}}{2}\right)}{t+\left(\frac{\sqrt{3}}{2}\right)}\right|+c$

$I=\frac{1}{\sqrt{3}} \ln \left|\frac{\sin x-\left(\frac{\sqrt{8}}{2}\right)}{\sin x+\left(\frac{\sqrt{8}}{2}\right)}\right|+c$