$$
\begin{array}{l}
\text { Let } I=\int_{0}^{\pi} \frac{x d x}{1+\cos \alpha \cdot \sin x} \quad \text { .. } \\
\Rightarrow \quad I=\int_{0}^{\pi} \frac{(\pi-x)}{1+\cos \alpha \cdot \sin (\pi-x)} d x \quad \ldots
\end{array}
$$
On adding Eqs. (i) and (ii), we get
$$
\begin{aligned}
2 I=\pi \int_{0}^{\pi} \frac{d x}{1+\cos \alpha \cdot \sin x} \\
=\pi \int_{0}^{\pi} \frac{d x}{1+\cos \alpha\left(\frac{2 \tan x / 2}{1+\tan ^{2} x / 2}\right)} \\
=\pi \int_{0}^{\pi} \frac{\sec ^{2} x / 2 d x}{\left(1+\tan ^{2} x / 2\right)+\cos \alpha(2 \tan x / 2)}
\end{aligned}
$$
Put $\quad \tan x / 2=t$
$$
\Rightarrow \quad(1 / 2) \sec ^{2} x / 2 d x=d t
$$
$\therefore$
$$
\begin{array}{l}
2 I=\pi \int_{0}^{\infty} \frac{2 d t}{1+t^{2}+2 t \cos \alpha} \\
I=\pi \int_{0}^{\infty} \frac{d t}{1+t^{2}+2 t \cos \alpha} \\
=\pi \int_{0}^{\infty} \frac{d t}{(t+\cos \alpha)^{2}+\sin ^{2} \alpha} \\
\quad=\frac{\pi}{\sin \alpha}\left[\tan ^{-1}\left(\frac{t+\cos \alpha}{\sin \alpha}\right)\right]_{0}^{\infty} \\
I=\frac{\pi \alpha}{\sin \alpha}
\end{array}
$$