Question

\[\int\limits_0^\pi \frac{x \sin x}{1 + \sin x} dx\]

Answer 1

\[Let I = \int_0^\pi \frac{x \sin x}{1 + \sin x} d x ................(1)\]
\[ = \int_0^\pi \frac{\left( \pi - x \right)\sin\left( \pi - x \right)}{1 + \sin\left( \pi - x \right)} dx\]
\[ = \int_0^\pi \frac{\left( \pi - x \right) \sin x}{1 + \sin x} d x ...................(2)\]
\[\text{Adding (1) and (2) we get} \]
\[2I = \int_0^\pi \left( x + \pi - x \right)\frac{\sin x}{1 + \sin x} d x \]
\[ = \int_0^\pi \frac{\pi \sin x}{1 + \sin x} d x\]
\[ = \pi \int_0^\pi \frac{1 + sinx - 1}{1 + sinx}dx\]
\[ = \pi \int_0^\pi dx - \pi \int_0^\pi \frac{1}{1 + sinx}dx\]
\[ = \pi \int_0^\pi dx - \pi \int_0^\pi \frac{\left( 1 - sinx \right)}{\left( 1 + sinx \right)\left( 1 - sinx \right)}dx\]
\[ = \pi \int_0^\pi dx - \pi \int_0^\pi \frac{\left( 1 - sinx \right)}{1 - \sin^2 x}dx\]
\[ = \pi \int_0^\pi dx - \pi \int_0^\pi \frac{\left( 1 - sinx \right)}{\cos^2 x}dx\]
\[ = \pi \int_0^\pi dx - \pi \int_0^\pi \left( \sec^2 x - \sec x \tan x \right)dx\]
\[ = \pi \left[ x \right]_0^\pi - \pi \left[ tanx - secx \right]_0^\pi \]
\[ = \pi^2 - \pi\left( 0 + 1 - 0 + 1 \right)\]
\[ = \pi^2 - 2\pi\]
\[Hence\ I = \pi\left( \frac{\pi}{2} - 1 \right)\]