Question

Evaluate :${\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{xSinx\cos x}{{\cos }^{4}x+{\sin }^{4}x}dx$

• A. $\frac{{\mathrm{\pi }}^2}{4}$
• B. $\frac{{\mathrm{\pi }}^2}{8}$
• C. $\frac{{\mathrm{\pi }}^2}{16}$
• D. $\frac{{\mathrm{\pi }}^2}{32}$

Answer 1

The correct option is D

$\frac{{\mathrm{\pi }}^2}{32}$

Evaluate the given integral :

Step 1: Re-expressing the integral

Let the integral be : $I={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{xSinx\cos x}{{\cos }^{4}x+{\sin }^{4}x}dx\to \left(i\right)$

substitute : $x=\left(0+\frac{\mathrm{\pi }}{2}-x\right)$ in equation $\left(1\right)$

$\begin{array}{rcl}I& =& {\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\left(0+\frac{\mathrm{\pi }}{2}-x\right)Sin\left(0+\frac{\mathrm{\pi }}{2}-x\right)\cos \left(0+\frac{\mathrm{\pi }}{2}-x\right)}{{\cos }^4\left(0+\frac{\mathrm{\pi }}{2}-x\right)+{\sin }^4\left(0+\frac{\mathrm{\pi }}{2}-x\right)}dx\\ & =& {\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\left(\frac{\mathrm{\pi }}{2}-x\right)\cos xSinx}{{\sin }^{4}x+{\cos }^{4}x}dx\to \left(ii\right)\left[\because \sin \left(\frac{\mathrm{\pi }}{2}-x\right)=\cos x,\cos \left(\frac{\mathrm{\pi }}{2}-x\right)=\sin x\right]\\ & & \end{array}$

Step 2: Add $\left(i\right)$ and $\left(ii\right)$

$\begin{array}{rcl}2I& =& {\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{xSinx\cos x}{{\cos }^{4}x+{\sin }^{4}x}dx+{\int }_0^{\frac{\mathrm{\pi }}{2}}\left(\frac{\mathrm{\pi }}{2}-x\right)\frac{\cos xSinx}{{\sin }^{4}x+{\cos }^{4}x}dx\\ & =& {\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\mathrm{\pi }}{2}\frac{Sinx\cos x}{{\cos }^{4}x+{\sin }^{4}x}dx\\ & =& \frac{\mathrm{\pi }}{2}{\int }_0^{\frac{\mathrm{\pi }}{2}}{\frac{Sinx\cos x}{{\left({\cos }^{2}x+{\sin }^{2}x\right)}^2-2{\cos }^{2}x{\sin }^{2}x}}^{}dx\left[\because (a^2+b^2)={(a+b)}^2-2ab\right]\\ & =& \frac{\mathrm{\pi }}{2}{\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{Sinx\cos x}{1-2\left(1-{\sin }^{2}x\right){\sin }^{2}x}dx\left[\because {\sin }^{2}x+{\cos }^{2}x=1\right]\end{array}$

Substitute $t={\sin }^{2}x$

$\therefore dt=2\sin x\cos x$

finding the Limits

$$\begin{gathered} x=0\Rightarrow t={\sin }^2\left(0\right)=0 \\ x=\frac{\mathrm{\pi }}{2}\Rightarrow t={\sin }^2\left(\frac{\mathrm{\pi }}{2}\right)=1 \end{gathered}$$

$\begin{array}{rcl}2I& =& \frac{\mathrm{\pi }}{2}{\int }_0^1\frac{1}{2}\frac{dt}{1-2t\left(1-t\right)}\\ & =& \frac{\mathrm{\pi }}{4}{\int }_0^1\frac{dt}{2t^2-2t+1}\\ & =& \frac{\mathrm{\pi }}{8}{\int }_0^1\frac{dt}{t^2-t+\frac{1}{2}+\frac{1}{4}-\frac{1}{4}}\left[addandsubtract\frac{1}{4}indenominator\right]\\ & =& \frac{\mathrm{\pi }}{8}{\int }_0^1\frac{dt}{{\left(t-{\displaystyle \frac{1}{2}}\right)}^2+{\left(\frac{1}{2}\right)}^2}\left[\because (a^2+b^2)={(a+b)}^2-2ab\right]\\ & =& \frac{\mathrm{\pi }}{8}{\left[{\tan }^{-1}\frac{t-\frac{1}{2}}{\frac{1}{2}}\right]}_0^1\left[\because \int \frac{dx}{1+x^2}={\tan }^{-1}x\right]\\ & =& \frac{\mathrm{\pi }}{8}{\left[{\tan }^{-1}\left(2t-1\right)\right]}_0^1\\ & =& \frac{\mathrm{\pi }}{8}\left[{\tan }^{-1}\left(2.1-1\right)-{\tan }^{-1}\left(2.0-1\right)\right]\\ & =& \frac{\mathrm{\pi }}{8}\left[{\tan }^{-1}\left(1\right)-{\tan }^{-1}\left(-1\right)\right]\\ & =& \frac{\mathrm{\pi }}{8}\left[\frac{\mathrm{\pi }}{4}-\left(\frac{\mathrm{\pi }}{4}\right)\right]\\ & =& \frac{{\mathrm{\pi }}^2}{16}\\ I& =& \frac{{\mathrm{\pi }}^2}{16}\end{array}$

Hence, option (D) is the correct answer.