Evaluate: _0^ /2 x sin x cos x sin^ 4 x + cos^ 4 x dx. - Vidyadip

Question

Evaluate: $\int\limits_0^{\pi/2}\frac{\text{x sin x cos x}}{\text{sin}^{4}\text{x + cos}^{4}\text{x}}\text{dx}$.

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Answer

$\text{I}=\int\limits_{0}^{\pi/2}\frac{\text{x sin x cos x}}{\text{sin}^{4}\text{x}+\text{cos}^{4}\text{x}}\text{dx}\ .....\text{(i)}$
$\text{I}=\int\limits_{0}^{\frac{\pi}{2}}{\pi}\frac{(\frac{\pi}{2}-x)\sin(\frac{\pi}{2}-x)\cos(\frac{\pi}{2}-x)}{\sin^{4}(\frac{\pi}{2}-x)+\cos^{4}(\frac{\pi}{2}-x)}\text{dx}=\int\limits_0^{\frac{\pi}{2}}\frac{(\frac{\pi}{2}-x)\text{cos x sin x dx}}{{\sin x}^4+\cos x^{4}}\ ......\text{(ii)}$
$2I = \text{I}=\frac{\pi}{2}\int\limits_{0}^{\frac{\pi}{2}}\frac{\text{sin x cos x}}{\text{sin}^{4}+\text{cos}^{4}\text{x}}\text{dx}=\frac{\pi}{2}\int\limits_0^{\frac{\pi}{2}}\frac{\text{tan x}\cdot\text{sec}^{2}\text{x dx}}{\text{1+tan}^{4}\text{x}}$
$\text{I}=\frac{\pi}{4}\frac{1}{2}\int\limits_0^{\infty}\frac{\text{dt}}{\text{1+t}^{2}}$ where $t = \tan^2 x.$
$=\frac{\pi}{8}[\tan ^{-1}\text{t}]_{0}^\infty=\frac{\pi}{8}\cdot\frac{\pi}{2}=\frac{\pi^{2}}{16}$.

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