Question

Evaluate $\int \frac{dx}{5+4\cos x}$

• A. $\frac{2}{3}{\tan }^{-1}\left(\frac{1}{3}\tan x\right)+c$
• B. $\frac{2}{3}{\tan }^{-1}\left(\frac{1}{3}\tan \frac{x}{2}\right)+c$
• C. $\frac{1}{3}{\tan }^{-1}\left(\frac{1}{3}\tan x\right)+c$
• D. $\frac{1}{3}{\tan }^{-1}\left(\frac{1}{3}\tan \frac{x}{2}\right)+c$

Answer 1

The correct option is B $\frac{2}{3}{\tan }^{-1}\left(\frac{1}{3}\tan \frac{x}{2}\right)+c$

Given,

$\int \frac{dx}{5+4\cos \left(x\right)}$

substitute $u=\tan \left(\frac{x}{2}\right)$

$=\int \frac{2}{u^2+9}du$

put $u=3v$

$=2\cdot \int \frac{1}{3(v^2+1)}dv$

$=\frac{2}{3}{\tan }^{-1}v$

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

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