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

and \[ I = \int_0^\pi {(\pi – x)\sin (\pi – x){{\cos }^2}(\pi – x)} dx\]

\[ \Rightarrow \] \[ I = \int_0^\pi {(\pi – x)} \sin x{\cos ^2}xdx\] ……(ii)

Adding Eqs. (i) and (ii), we get

\[ 2I = \int_0^\pi \pi \sin x{\cos ^2}xdx\]

Let’s put \[ \cos x = t\]

\[ \Rightarrow \] \[ – \sin xdx = dt\]

As \[ x \to 0,\] then \[ t \to 1\]

and \[x \to \pi ,\] then \[ t \to – 1\]

therefore,\[I = – \pi \int_1^{ – 1} {{t^2}} dt \Rightarrow I = – \pi \left[ {\frac{{{t^3}}}{3}} \right]_1^{ – 1}\]

\[ \Rightarrow \] \[2I = – \frac{\pi }{3}[ – 1 – 1] \Rightarrow 2I = \frac{{2\pi }}{3}\]

 

therefore,\[I = \frac{\pi }{3}\]

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