Explanation

(c)
Let $I = \int_{0}^{\pi}\frac{x\ dx}{1 + \sin x}\$ …(i)
$\Rightarrow I = \int_{0}^{\pi}{\frac{(\pi - x)dx}{1 + \sin x}\ \ \ldots(ii)}$
On adding Eqs. (i) and (ii), we get
$2I = \int_{0}^{\pi}\frac{\pi\ dx}{1 + \sin x}$
$= \pi\int_{0}^{\pi}{\frac{1 - \sin x}{1 - \sin^{2}x}dx}$
$\Rightarrow \ 2I = \pi\int_{0}^{\pi}{(\sec^{2}{x - \sec{x\tan{x)dx}}}}$
$\Rightarrow \ 2I = \pi\left\lbrack \tan\ x - sec\ x \right\rbrack_{0}^{\pi} = 2\pi$
$\Rightarrow \ I = \pi$