Question

Integrate with respect to $x$: $\frac{1}{2+\cos x}$

Answer 1

Consider the following integral .

$I=\underset{}{\overset{}{\int }}\frac{1}{2+\cos x}dx$

Using a given formula. and differentiate both side w.r.t $x$

$I=\underset{}{\overset{}{\int }}\frac{1}{2+\cos x}dx$

$u=\tan \frac{x}{2}$

$\cos x=\frac{1-u^2}{1+u^2}$

$dx=\frac{2du}{1+u^2}$

$=\underset{}{\overset{}{\int }}\frac{1}{2+\frac{1-u^2}{1+u^2}}\frac{2du}{1+u^2}$

$=\underset{}{\overset{}{\int }}\frac{2du}{3+u^2}$

$=2\underset{}{\overset{}{\int }}\frac{du}{{\left(\sqrt{3}\right)}^2+u^2}$

$I=\frac{2}{3}{\tan }^{-1}\left(\frac{u}{\sqrt{3}}\right)+C$

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

Hence, this is the correct answer.