Question

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

Answer 1

$\int \frac{x^2}{x\sin x+\cos x}dx$

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

$du=\frac{{\sec }^{2}x}{2}dx$

$\cos x=\frac{1-{\tan }^2\frac{x}{2}}{{\sec }^2\frac{x}{2}}$

$\sin x=\frac{2\tan \frac{x}{2}}{{\sec }^2\frac{x}{2}}$

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

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

$=2(-\int \frac{1}{u^2+1}du+\int 1du)$

$=2(-{\tan }^{-1}u+u)$

Re-substituting $u=\tan \frac{x}{2}$

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

$\therefore \int \frac{x^2}{x\sin x+\cos x}dx=2[-{\tan }^{-1}\left(\tan \frac{x}{2}\right)+\tan \frac{x}{2}]$