In integrals like this the following substitution works 'almost always'...
$\displaystyle \displaystyle t= \tan \frac{x}{2} \implies dx= 2\ \frac{dt}{1+t^{2}}, \sin x = \frac{2t}{1+t^{2}}, \cos x = \frac{1-t^{2}}{1+t^{2}}$
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$