#### Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise  Question 73 Maths Textbook Solution.

$2\left(\frac{1}{\sqrt{3}}\right) \tan ^{-1} \frac{\left(\tan \left(\frac{x}{2}\right)\right)}{\sqrt{3}}$

Hint:

To solve the given statement we will write $\cos x \operatorname{as} \frac{1-\tan ^{2}\left(\frac{x}{2}\right)}{1+\tan ^{2}\left(\frac{x}{2}\right)}$

Given:

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

Solution:

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

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

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

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

$=2\left(\frac{1}{\sqrt{3}}\right) \tan ^{-1} \frac{\left(\tan \left(\frac{\mathrm{x}}{2}\right)\right)}{\sqrt{3}}$