By letting $\displaystyle x=cos\theta$, differentiating to get $\displaystyle \frac{dx}{d\theta}=-sin\theta$, which becomes $\displaystyle dx=-sin\theta{d\theta}$
Substituting into the integral, you get
$\displaystyle \frac{-sin\theta{d\theta}}{(1+cos^2\theta)(\sqrt{1-cos^2\theta})}$
However $\displaystyle 1-cos^2\theta=sin^2\theta
$
So simplifying the expression
$\displaystyle \frac{-d\theta{sin\theta}}{(1+cos^2\theta)(\sqrt{sin^2\th eta})}$
$\displaystyle \frac{-d\theta}{1+cos^2\theta}$
And as we've changed the variable, we need to change the limits of the integral
As $\displaystyle x=cos\theta$, when$\displaystyle x=1,$ $\displaystyle \theta=0 $
And when $\displaystyle x=-1, \theta=\pi$
So the integral becomes
$\displaystyle 2\int_{\pi}^{0} \frac{-d\theta}{1+cos^2\theta}$
Becoming
$\displaystyle -2\int_{\pi}^{0} \frac{d\theta}{1+cos^2\theta}$