Question

${\int }_0^{\frac{\mathrm{\pi }}{2}}x{\sin }^{2}x{\cos }^{2}xdx=$

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

Answer 1

The correct option is D

None of these

Explanation for Correct answer:

Finding the value of the given integral:

Let $I={\int }_0^{\frac{\mathrm{\pi }}{2}}x{\sin }^{2}x{\cos }^{2}xdx\to \left(i\right)$

We know that, ${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f\left(a+b-x\right)dx$

Replace $x\to 0+\frac{\mathrm{\pi }}{2}-x\left[\text{where}a=0,b=\frac{\mathrm{\pi }}{2}\right]$

$$\begin{gathered} I={\int }_0^{\frac{\mathrm{\pi }}{2}}\left(0+\frac{\mathrm{\pi }}{2}-x\right){\sin }^2\left(0+\frac{\mathrm{\pi }}{2}-x\right){\cos }^2\left(0+\frac{\mathrm{\pi }}{2}-x\right)dx \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\left(\frac{\mathrm{\pi }}{2}-x\right){\cos }^{2}x{\sin }^{2}xdx\to \left(ii\right)\left[\because \sin \left(\frac{\mathrm{\pi }}{2}-x\right)=\cos x,\cos \left(\frac{\mathrm{\pi }}{2}-x\right)=\sin x\right] \end{gathered}$$

Adding equation $\left(i\right)$ and $\left(ii\right)$, we get

$\begin{array}{rcl}2I& =& {\int }_0^{\frac{\mathrm{\pi }}{2}}x{\sin }^{2}x{\cos }^{2}xdx+{\int }_0^{\frac{\mathrm{\pi }}{2}}\left(\frac{\mathrm{\pi }}{2}-x\right){\cos }^{2}x{\sin }^{2}xdx\\ & =& {\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\mathrm{\pi }}{2}{\cos }^{2}x{\sin }^{2}xdx\\ & =& \frac{\mathrm{\pi }}{2}{\int }_0^{\frac{\mathrm{\pi }}{2}}{\left(\frac{\sin 2x}{2}\right)}^{2}dx\left[\because \sin 2x=2\sin x\cos x\right]\\ & =& \frac{\mathrm{\pi }}{8}{\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{1-\cos 4x}{2}dx\left[\because \cos 2x=1-2{\sin }^{2}x\right]\\ & =& \frac{\mathrm{\pi }}{16}{\left[x-\frac{\sin 4x}{4}\right]}_0^{\frac{\mathrm{\pi }}{2}}\left[\because \int \cos xdx=\sin x\right]\\ & =& \frac{\mathrm{\pi }}{16}\times \frac{\mathrm{\pi }}{2}\left[\because \sin 2\mathrm{\pi }=0\right]\\ & =& \frac{{\mathrm{\pi }}^2}{32}\end{array}$

$\begin{array}{rcl}2I& =& \frac{{\mathrm{\pi }}^2}{32}\\ & \Rightarrow & I=\frac{{\mathrm{\pi }}^2}{64}\end{array}$

Hence, option (D) is the correct answer.