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प्रश्न
\[\int\limits_0^\pi \frac{x \sin x}{1 + \cos^2 x} dx\]
उत्तर
\[Let, I = \int_0^\pi \frac{x \sin x}{1 + \cos^2 x} d x ...........(1)\]
\[ = \int_0^\pi \frac{\left( \pi - x \right) \sin\left( \pi - x \right)}{1 + \cos^2 \left( \pi - x \right)} d x\]
\[ = \int_0^\pi \frac{\left( \pi - x \right) \sin x}{1 + \cos^2 x} d x ................(2)\]
Adding (1) and (2)
\[2I = \int_0^\pi \left[ \frac{x \sin x}{1 + \cos^2 x} + \frac{\left( \pi - x \right) \sin x}{1 + \cos^2 x} \right] d x\]
\[ = \int_0^\pi \frac{\pi \sin x}{1 + \cos^2 x} d x \]
\[ = \pi \left[ - \tan^{- 1} \left( cosx \right) \right]_0^\pi \]
\[ = - \pi\left[ \tan^{- 1} \left( - 1 \right) - \tan^{- 1} \left( 1 \right) \right]\]
\[ = - \pi\left( - \frac{\pi}{4} - \frac{\pi}{4} \right)\]
\[ = \frac{\pi^2}{2}\]
\[Hence, I = \frac{\pi^2}{4}\]
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