Question

${\int }_0^{\pi /2}x{\sin }^{2}x{\cos }^{2}xdx$ is equal to

• A. $\frac{{\pi }^2}{32}$
• B. $\frac{{\pi }^2}{16}$
• C. $\frac{\pi }{32}$
• D. none of these

Answer 1

Answer: (A)

Let $I={\int }_0^{\pi /2}x{\sin }^{2}x{\cos }^{2}xdx$ ...(i) $I={\int }_0^{\pi /2}\left(\frac{\pi }{2}-x\right){\sin }^{2}x{\cos }^{2}xdx$ ...(ii) Adding Eqs. (i) and (ii) we get $2I=\frac{\pi }{2}{\int }_0^{\pi /2}{\sin }^{2}x{\cos }^{2}xdx$ $=\frac{\pi }{8}{\int }_0^{\pi /2}{\sin }^{2}2xdx$ $\Rightarrow$ $2I=\frac{\pi }{8}{\left[x-\frac{\sin 4x}{4}\right]}_0^{\pi /2}=\frac{\pi }{8}\left[\frac{\pi }{2}\right]$ $\Rightarrow$ $I=\frac{{\pi }^2}{32}$