$$\begin{gathered} \mathrm{Let} \mathrm{I} = {\int }_0^{\mathrm{\pi }} \frac{\mathrm{x}}{1 + \mathrm{sinx}} \mathrm{dx} .........\left(\mathrm{i}\right) \\ \mathrm{Using} \mathrm{the} \mathrm{property} {\int }_0^{\mathrm{a}} \mathrm{f} ( \mathrm{x} ) \mathrm{dx} = {\int }_0^{\mathrm{a}} \mathrm{f} ( \mathrm{a} -\mathrm{x} ) \mathrm{dx} \\ \Rightarrow \mathrm{I} = {\int }_0^{\mathrm{\pi }} \frac{\mathrm{\pi } - \mathrm{x}}{1 + \sin \left( \mathrm{\pi } - \mathrm{x} \right)} \mathrm{dx} \\ = {\int }_0^{\mathrm{\pi }} \frac{\mathrm{\pi } - \mathrm{x}}{1 + \sin \mathrm{x}} \mathrm{dx} .........\left(\mathrm{ii}\right) \end{gathered}$$

Now adding (i) and (ii), we get

$$\begin{gathered} 2\mathrm{I} = {\int }_0^{\mathrm{\pi }} \frac{\mathrm{x}}{1 + \mathrm{sinx}} \mathrm{dx} + {\int }_0^{\mathrm{\pi }} \frac{\mathrm{\pi } - \mathrm{x}}{1 + \mathrm{sinx}} \mathrm{dx} \\ = {\int }_0^{\mathrm{\pi }} \frac{\mathrm{\pi }}{1 + \mathrm{sinx}} \mathrm{dx} \\ = \mathrm{\pi } {\int }_0^{\mathrm{\pi }} \frac{1}{1 + \mathrm{sinx}} \mathrm{dx} \\ = \mathrm{\pi } {\int }_0^{\mathrm{\pi }} \frac{\left( 1 - \mathrm{sinx} \right)}{\left( 1 - {\sin }^2\mathrm{x} \right)} \mathrm{dx} \\ = \mathrm{\pi } {\int }_0^{\mathrm{\pi }} \frac{\left( 1 - \mathrm{sinx} \right)}{\left( {\cos }^2\mathrm{x} \right)} \mathrm{dx} \\ = \mathrm{\pi } \left[ {\int }_0^{\mathrm{\pi }} \left(\frac{1}{{\cos }^2\mathrm{x}} - \frac{\mathrm{sinx}}{{\cos }^2\mathrm{x}} \right) \mathrm{dx} \right] \end{gathered}$$

$$\begin{gathered} = \mathrm{\pi } \left[ {\int }_0^{\mathrm{\pi }} {\sec }^2 \mathrm{x} - \mathrm{secx} \mathrm{tanx} \mathrm{dx}\right] \\ = \mathrm{\pi } \left[{\int }_0^{\mathrm{\pi }} {\sec }^2 \mathrm{x} \mathrm{dx} - {\int }_0^{\mathrm{\pi }} \mathrm{secx} \mathrm{tanx} \mathrm{dx} \right] \\ = \mathrm{\pi } \left( {\left[ \mathrm{tanx} \right]}_0^{\mathrm{\pi }} - {\left[ \mathrm{secx} \right]}_0^{\mathrm{\pi }} \right) \\ \Rightarrow 2\mathrm{I} = \mathrm{\pi } \left( 2 \right) \\ \Rightarrow \mathrm{I} = \mathrm{\pi } \\ \mathrm{So}, {\int }_0^{\mathrm{\pi }} \frac{\mathrm{x}}{1 + \mathrm{sinx}} \mathrm{dx} = \mathrm{\pi } \end{gathered}$$