Question

Solve : $\int \frac{dx}{2-3\cos 2x}$

Answer 1

$\int \frac{dx}{2-3\cos 2x}dx$

$=-\int \frac{dx}{3\cos 2x-2}dx$

substitute $u=2x\Rightarrow du=2dx$

$=-\int \frac{dx}{3\frac{1-{\tan }^{\frac{u}{2}}}{{\tan }^{\frac{u}{2}}+1}\cos 2x-2}dx$

substitute $v={\tan }^{\frac{u}{2}}\to du=\frac{2}{{\sec }^2\frac{u}{2}}dv=\frac{2}{v^2+1}dv$

$=-2\int \frac{1}{5v^2-1}dv$

$=\frac{\ln \left(|5\tan x-\sqrt{5}|\right)}{2\sqrt{5}}-\frac{\ln \left(|5\tan x+\sqrt{5}|\right)}{2\sqrt{5}}+C$

$=\frac{\ln \left(|5\tan x-\sqrt{5}|\right)-\ln \left(|5\tan x+\sqrt{5}|\right)}{2\sqrt{5}}+C$