Question

\[\int\limits_0^\pi \frac{x \sin x}{1 + \cos^2 x} dx\]

Answer 1

\[Let, I = \int_0^\pi \frac{x \sin x}{1 + \cos^2 x} d x ...........(1)\]

\[ = \int_0^\pi \frac{\left( \pi - x \right) \sin\left( \pi - x \right)}{1 + \cos^2 \left( \pi - x \right)} d x\]

\[ = \int_0^\pi \frac{\left( \pi - x \right) \sin x}{1 + \cos^2 x} d x ................(2)\]

Adding (1) and (2)

\[2I = \int_0^\pi \left[ \frac{x \sin x}{1 + \cos^2 x} + \frac{\left( \pi - x \right) \sin x}{1 + \cos^2 x} \right] d x\]

\[ = \int_0^\pi \frac{\pi \sin x}{1 + \cos^2 x} d x \]

\[ = \pi \left[ - \tan^{- 1} \left( cosx \right) \right]_0^\pi \]

\[ = - \pi\left[ \tan^{- 1} \left( - 1 \right) - \tan^{- 1} \left( 1 \right) \right]\]

\[ = - \pi\left( - \frac{\pi}{4} - \frac{\pi}{4} \right)\]

\[ = \frac{\pi^2}{2}\]

\[Hence, I = \frac{\pi^2}{4}\]