Question

Solve : $\int \frac{\sin 2x}{(a+b\cos x{)}^2}dx$

Answer 1

Given : $\int \frac{\sin 2x}{(a+b\cos x{)}^2}$

$\int \frac{\sin 2x}{(a+b\cos x{)}^2}=\int \frac{2\sin x\cos xdx}{(a+b\cos x{)}^2}$

Let : $t=\cos x$

$-dt=\sin xdx\int \frac{-2tdt}{(a+bt{)}^2}=\frac{A}{(a+bt)}+\frac{B}{(a+bt{)}^2}\frac{-2t}{(a+bt{)}^2}=\frac{A(a+bt)+B}{(a+bt{)}^2}$

Comparing coefficient of $t$

$-2=Ab0=aA+BA=-\frac{2}{b}B=-aA=+\left(\frac{2a}{b}\right)\int \frac{-2tdt}{(a+bt{)}^2}=\int \frac{-2dt}{b(a+bt)}+\frac{2a}{b}\int \frac{1}{(a+bt{)}^2}=-\frac{2}{b^2}\ln (a+bt)+\frac{2a}{b^2}[-\frac{1}{(a+bt)}]=-\frac{2}{b^2}\ln (a+b\cos x)+\frac{2a}{b^2}[-\frac{1}{(a+b\cos x)}]+C$