Question

\[\int\limits_0^\pi x \cos^2 x\ dx\]

Answer 1

\[Let\ I = \int_0^\pi x \cos^2 x\ d\ x . . . (i) \]
\[ = \int_0^\pi \left( \pi - x \right) \cos^2 \left( \pi - x \right)\ d\ x\]
\[ = \int_0^\pi \left( \pi - x \right) \cos^2 x\ dx . . . (ii)\]
\[\text{Adding (i) and (ii) we get}\]
\[2I = \int_0^\pi \left( x + \pi - x \right) \cos^2 x\ dx\]
\[ = \int_0^\pi \pi \cos^2 x\ dx\]
\[ = \pi \int_0^\pi \frac{1 + \cos2x}{2} dx\]
\[ = \frac{\pi}{2} \int_0^\pi \left( 1 + \cos2x \right) dx\]
\[ = \frac{\pi}{2} \left[ x + \frac{\sin2x}{2} \right]_0^\pi \]
\[ = \frac{\pi}{2}\left( \pi - 0 \right)\]
\[ Hence\ I = \frac{\pi^2}{4}\]