Explanation

(b)
Let $I = \int_{0}^{\pi}{\frac{1}{1 + \sin x}dx}$
$= \int_{0}^{\pi}{\frac{1}{1 + \frac{2\ tan\frac{x}{2}}{1 + \tan^{2}\frac{x}{2}}}dx}$
$= \int_{0}^{\pi}{\frac{\sec^{2}\frac{x}{2}}{\left( 1 + tan\frac{x}{2} \right)^{2}}dx}$
Put $\tan\frac{x}{2} = t \Rightarrow \frac{1}{2}\sec^{2}{\frac{x}{2}dx = dt}$
$\therefore I = \int_{0}^{\infty}{\frac{2dt}{(1 + t)^{2}} = \left\lbrack - \frac{2}{1 + t} \right\rbrack_{0}^{\infty}} = 2$