$\int\limits_{0}^{\pi} \frac{x}{1+\sin x}$ dx Using the property $\int\limits_{0}^{a}$ f (x) dx = $\int\limits_{0}^{a}$ f (a - x) dx ⇒ I = $\int\limits_{0}^{\pi} \frac{\pi - x}{1+\sin(\pi - x)}$ dx = $\int\limits_{0}^{\pi}$ $\frac{\pi - x}{1+\sin x}$ dx ... (2) Now adding (1) and (2), we get 2I = $\int\limits_{0}^{\pi}$ $\frac{x}{1+\sin x}$ dx + $\int\limits_{0}^{\pi}$ $\frac{\pi - x}{1+\sin x}$ dx = $\int\limits_{0}^{\pi}$ $\frac{\pi}{1+\sin x}$ dx = π $\int\limits_{0}^{\pi}$ $\frac{1}{1+\sin x}$ dx = π $\int\limits_{0}^{\pi}$ $\frac{1-\sin x}{1-\sin^2 x}$ dx == π $\int\limits_{0}^{\pi}$ $\frac{1-\sin x}{\cos^2 x}$ dx = π $\left[ \int\limits_{0}^{\pi} \left( \frac{1}{\cos^2 x} - \frac{\sin x}{\cos^2 x} \right) dx \right]$ = π $\left[ \int\limits_{0}^{\pi} \sec^2 x - \sec x \tan x \, dx \right]$ = π $\left[ \int\limits_{0}^{\pi} \sec^2 x \, dx - \int\limits_{0}^{\pi} \sec x \tan x \, dx \right]$ = π $\left( [\tan x]_{0}^{\pi} - [\sec x]_{0}^{\pi} \right)$ ⇒ 2I = π (2) ⇒ I = π So, $\int\limits_{0}^{\pi} \frac{x}{1+\sin x}$ dx = π