Evaluate: ^ /2 _ 0 x x x ^4x+ ^4x dx

Question

Evaluate: $\int\limits^{\pi/2}_{0}\frac{x\sin x\cos x}{\sin^4x+\cos^4x}\text{d}x$

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Answer

$\text{let I}=\int\limits^{\pi/2}_{0}\frac{x\sin x\cos x}{\sin^4x+\cos^4x}\text{d}x;$ $\therefore\ \ \text{I}=\int\limits^{\pi/2}_{0}\frac{\big(\pi/2-x\big)\cos x\sin x}{\cos^4x+\sin^4x}\text{d}x$
Adding we get, $\text{2I}=\frac{\pi}{2}\int\limits^{\pi/2}_{0}\frac{\sin x\cos x}{\sin^4x+\cos^4x}\text{d}x;$ $\ \ =\int\limits^{\pi/2}_{0}\frac{2\tan x \sec^2 x}{1+(\tan^2x)^2}\text{d}x$
$\ \ =\frac{\pi}{4}\tan^{-1}(\tan^2x)\bigg]^{\pi/2}_0=\frac{\pi^2}{8}$
$\therefore\ \ \text{I}=\frac{\pi^2}{16}$

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