Question

Integrate the following $\int \frac{x^2}{{(x\sin x+\cos x)}^2}dx$

Answer 1

Put $x=\tan \theta$
$I=\int \frac{ta{n}^2\theta se{c}^2\theta }{{\left(\frac{sin\theta }{cos\theta }sin\left(tan\theta \right)+cos\left(tan\theta \right)\right)}^2}d\theta$
$=\int \frac{ta{n}^2\theta }{co{s}^2(tan\theta -\theta )}d\theta$
Put $tan\theta -\theta =u\Rightarrow ta{n}^2\theta d\theta =du$
$I=\int \frac{du}{co{s}^{2}u}=\int se{c}^{2}udu=tan(tan\theta -\theta )$
$=tan(x-ta{n}^{-1}x)=\frac{tanx-x}{1+xtanx}$
$=\frac{sinx-xcosx}{cosx+x\sin x}+C$