Question

# Evaluate :${\int }_{0}^{\frac{\mathrm{\pi }}{2}}\frac{xSinx\mathrm{cos}x}{{\mathrm{cos}}^{4}x+{\mathrm{sin}}^{4}x}dx$

A

$\frac{{\mathrm{\pi }}^{2}}{4}$

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B

$\frac{{\mathrm{\pi }}^{2}}{8}$

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C

$\frac{{\mathrm{\pi }}^{2}}{16}$

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D

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

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Solution

## The correct option is D $\frac{{\mathrm{\pi }}^{2}}{32}$Evaluate the given integral :Step 1: Re-expressing the integralLet the integral be : $I={\int }_{0}^{\frac{\mathrm{\pi }}{2}}\frac{xSinx\mathrm{cos}x}{{\mathrm{cos}}^{4}x+{\mathrm{sin}}^{4}x}dx\to \left(i\right)$substitute : $x=\left(0+\frac{\mathrm{\pi }}{2}-x\right)$ in equation $\left(1\right)$$\begin{array}{rcl}I& =& {\int }_{0}^{\frac{\mathrm{\pi }}{2}}\frac{\left(0+\frac{\mathrm{\pi }}{2}-x\right)Sin\left(0+\frac{\mathrm{\pi }}{2}-x\right)\mathrm{cos}\left(0+\frac{\mathrm{\pi }}{2}-x\right)}{{\mathrm{cos}}^{4}\left(0+\frac{\mathrm{\pi }}{2}-x\right)+{\mathrm{sin}}^{4}\left(0+\frac{\mathrm{\pi }}{2}-x\right)}dx\\ & =& {\int }_{0}^{\frac{\mathrm{\pi }}{2}}\frac{\left(\frac{\mathrm{\pi }}{2}-x\right)\mathrm{cos}xSinx}{{\mathrm{sin}}^{4}x+{\mathrm{cos}}^{4}x}dx\to \left(ii\right)\left[\because \mathrm{sin}\left(\frac{\mathrm{\pi }}{2}-x\right)=\mathrm{cos}x,\mathrm{cos}\left(\frac{\mathrm{\pi }}{2}-x\right)=\mathrm{sin}x\right]\\ & & \end{array}$Step 2: Add $\left(i\right)$ and $\left(ii\right)$ $\begin{array}{rcl}2I& =& {\int }_{0}^{\frac{\mathrm{\pi }}{2}}\frac{xSinx\mathrm{cos}x}{{\mathrm{cos}}^{4}x+{\mathrm{sin}}^{4}x}dx+{\int }_{0}^{\frac{\mathrm{\pi }}{2}}\left(\frac{\mathrm{\pi }}{2}-x\right)\frac{\mathrm{cos}xSinx}{{\mathrm{sin}}^{4}x+{\mathrm{cos}}^{4}x}dx\\ & =& {\int }_{0}^{\frac{\mathrm{\pi }}{2}}\frac{\mathrm{\pi }}{2}\frac{Sinx\mathrm{cos}x}{{\mathrm{cos}}^{4}x+{\mathrm{sin}}^{4}x}dx\\ & =& \frac{\mathrm{\pi }}{2}{\int }_{0}^{\frac{\mathrm{\pi }}{2}}{\frac{Sinx\mathrm{cos}x}{{\left({\mathrm{cos}}^{2}x+{\mathrm{sin}}^{2}x\right)}^{2}-2{\mathrm{cos}}^{2}x{\mathrm{sin}}^{2}x}}^{}dx\left[\because \left({a}^{2}+{b}^{2}\right)={\left(a+b\right)}^{2}-2ab\right]\\ & =& \frac{\mathrm{\pi }}{2}{\int }_{0}^{\frac{\mathrm{\pi }}{2}}\frac{Sinx\mathrm{cos}x}{1-2\left(1-{\mathrm{sin}}^{2}x\right){\mathrm{sin}}^{2}x}dx\left[\because {\mathrm{sin}}^{2}x+{\mathrm{cos}}^{2}x=1\right]\end{array}$Substitute $t={\mathrm{sin}}^{2}x$$\therefore dt=2\mathrm{sin}x\mathrm{cos}x$finding the Limits $x=0⇒t={\mathrm{sin}}^{2}\left(0\right)=0\phantom{\rule{0ex}{0ex}}x=\frac{\mathrm{\pi }}{2}⇒t={\mathrm{sin}}^{2}\left(\frac{\mathrm{\pi }}{2}\right)=1$ $\begin{array}{rcl}2I& =& \frac{\mathrm{\pi }}{2}{\int }_{0}^{1}\frac{1}{2}\frac{dt}{1-2t\left(1-t\right)}\\ & =& \frac{\mathrm{\pi }}{4}{\int }_{0}^{1}\frac{dt}{2{t}^{2}-2t+1}\\ & =& \frac{\mathrm{\pi }}{8}{\int }_{0}^{1}\frac{dt}{{t}^{2}-t+\frac{1}{2}+\frac{1}{4}-\frac{1}{4}}\left[addandsubtract\frac{1}{4}indenominator\right]\\ & =& \frac{\mathrm{\pi }}{8}{\int }_{0}^{1}\frac{dt}{{\left(t-\frac{1}{2}\right)}^{2}+{\left(\frac{1}{2}\right)}^{2}}\left[\because \left({a}^{2}+{b}^{2}\right)={\left(a+b\right)}^{2}-2ab\right]\\ & =& \frac{\mathrm{\pi }}{8}{\left[{\mathrm{tan}}^{-1}\frac{t-\frac{1}{2}}{\frac{1}{2}}\right]}_{0}^{1}\left[\because \int \frac{dx}{1+{x}^{2}}={\mathrm{tan}}^{-1}x\right]\\ & =& \frac{\mathrm{\pi }}{8}{\left[{\mathrm{tan}}^{-1}\left(2t-1\right)\right]}_{0}^{1}\\ & =& \frac{\mathrm{\pi }}{8}\left[{\mathrm{tan}}^{-1}\left(2.1-1\right)-{\mathrm{tan}}^{-1}\left(2.0-1\right)\right]\\ & =& \frac{\mathrm{\pi }}{8}\left[{\mathrm{tan}}^{-1}\left(1\right)-{\mathrm{tan}}^{-1}\left(-1\right)\right]\\ & =& \frac{\mathrm{\pi }}{8}\left[\frac{\mathrm{\pi }}{4}-\left(\frac{\mathrm{\pi }}{4}\right)\right]\\ & =& \frac{{\mathrm{\pi }}^{2}}{16}\\ I& =& \frac{{\mathrm{\pi }}^{2}}{16}\end{array}$Hence, option (D) is the correct answer.

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