#### Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 64 Maths Textbook Solution.

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

Hint

We will write $\cos x \operatorname{as}\left(\frac{1-\tan ^{2} \frac{x \lambda}{2}}{1+\tan ^{2} \frac{x}{2}}\right)$

Given:

$\int \frac{1}{1+2 \cos x} d x$

Solution:

$I=\int \frac{1}{1+2 \cos x} d x$

$I=\int \frac{1}{1+2\left(\frac{1-\tan ^{2 \frac{x}{2}}}{1+\tan ^{2} \frac{x}{2}}\right)} d x$

$I=\int \frac{1+\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}+2-2 \tan ^{2} \frac{x}{2}} d x$

$I=\int \frac{\sec ^{2} \frac{x}{2}}{3-\tan ^{2} \frac{x}{2}} d x$

$I=\int \frac{2 d t}{3-t^{2}}$                                                $\left[\because \tan \frac{\mathrm{x}}{2}=t, d t=\sec ^{2}\left(\frac{x}{2}\right)\left(\frac{1}{2}\right) d x\right]$

$I=2 \int \frac{d t}{(\sqrt{3})^{2}-t^{2}}$                                $\left[\because \sec ^{2}\left(\frac{x}{2}\right) d x=2 d t\right]$

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

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

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