Question

Evaluate

$\int \frac{dx}{5-3cosx}$

Answer 1

$\begin{array}{c}{\int }^{}\frac{1}{5-3\cos \left(x\right)}dx=-{\int }^{}\frac{1}{3\cos x-5}dx\\ ={\int }^{}-\frac{{\sec }^2\left(\frac{x}{2}\right)}{2\left(4{\tan }^2\left(\frac{x}{2}\right)+1\right)}dx\\ substitute,u=2\tan \left(\frac{x}{2}\right)\\ =-\frac{1}{2}{\int }^{}\frac{1}{u^2+1}du\\ =-\frac{{\tan }^{-1}u}{2}\\ =\frac{{\tan }^{-1}\left(2\tan \left(\frac{x}{2}\right)\right)}{2}+C\end{array}$