Question

Evaluate the following integral:

\[\int_0^\pi x\sin x \cos^2 xdx\]

Answer 1

\[\text{Let I} =\int_0^\pi x\sin x \cos^2 xdx .....................(1)\]

Then,

\[I = \int_0^\pi \left( \pi - x \right)\sin\left( \pi - x \right) \cos^2 \left( \pi - x \right)dx ..................\left[ \int_0^a f\left( x \right)dx = \int_0^a f\left( a - x \right)dx \right]\]
\[ = \int_0^\pi \left( \pi - x \right)\sin x \cos^2 xdx .................(2)\]

Adding (1) and (2), we have

\[2I = \int_0^\pi \left( \pi - x + x \right)\sin x \cos^2 xdx\]
\[ \Rightarrow 2I = \pi \int_0^\pi \sin x \cos^2 xdx\]
\[ \Rightarrow 2I = - \pi \int_0^\pi \cos^2 x\left( - \sin x \right)dx\]
\[ \Rightarrow 2I = \left.- \pi \times \frac{\cos^3 x}{3}\right|_0^\pi .................\left[ \int \left[ f\left( x \right) \right]^n f'\left( x \right)dx = \frac{\left[ f\left( x \right) \right]^{n + 1}}{n + 1} + C \right]\]
\[ \Rightarrow 2I = - \frac{\pi}{3}\left( \cos^3 \pi - \cos^2 0 \right)\]

\[\Rightarrow 2I = - \frac{\pi}{3}\left( - 1 - 1 \right) = \frac{2\pi}{3}\]
\[ \Rightarrow I = \frac{\pi}{3}\]