245. Find $\int_{\pi / 2}^{\pi} \frac{2 \sin x d x}{(1-\cos x)^{2}}$.
Solution. Let's use the substitution $u=1-\cos x$, from which $d u=$ $=\sin x d x$. Then we will find the new limits of integration; substituting into the equation $u=1-\cos x$ the values $x_{1}=\pi / 2$ and $x_{2}=\pi$, we will respectively obtain $u_{1}=1-\cos (\pi / 2)=1$ and $u_{2}=1-\cos \pi=2$. The solution is written as follows:

$$
\begin{aligned}
& \int_{\pi / 2}^{\pi} \frac{2 \sin x d x}{(1-\cos x)^{2}}=\left|\begin{array}{l}
u=1-\cos x, u_{1}=1-\cos (\pi / 2)=1, \\
d u=\sin x d x ; u_{2}=1-\cos \pi=2
\end{array}\right|=2 \int_{u}^{u_{2}} \frac{d u}{u^{2}}=2 \int_{1}^{2} u^{-2} d u= \\
& =-\left.2 u^{-1}\right|_{1} ^{2}=-1+2=1
\end{aligned}
$$