Question

Evaluate $\int\limits_{0}^{\pi / 2} \frac{x \sin x \cos x}{\cos ^{4} x+\sin ^{4} x} d x$.

Answer 1

Let $I=\int\limits_{0}^{\pi / 2} \frac{x \sin x \cdot \cos x}{\cos ^{4} x+\sin ^{4} x} d x$
$\Rightarrow I=\int\limits_{0}^{\pi / 2} \frac{\left(\frac{\pi}{2}-x\right) \sin \left(\frac{\pi}{2}-x\right) \cdot \cos \left(\frac{\pi}{2}-x\right)}{\sin ^{4}\left(\frac{\pi}{2}-x\right)+\cos ^{4}\left(\frac{\pi}{2}-x\right)} d x$
$\Rightarrow I=\int\limits_{0}^{\pi / 2} \frac{\left(\frac{\pi}{2}-x\right) \cdot \sin x \cos x}{\cos ^{4} x+\sin ^{4} x} d x$
$\Rightarrow I \equiv \frac{\pi}{2} \int\limits_{0}^{\pi / 2} \frac{\sin x \cos x}{\sin ^{4} x+\cos ^{4} x} d x-\int\limits_{0}^{\pi / 2} \frac{x \sin x \cdot \cos x}{\sin ^{4} x+\cos ^{4} x} d x $
$=\frac{\pi}{2} \int\limits_{0}^{\pi / 2} \frac{\sin x \cdot \cos x}{\sin ^{4} x+\cos ^{4} x} d x-I$
$\Rightarrow 2 I=\frac{\pi}{2} \int\limits_{0}^{\pi / 2} \frac{\tan x \cdot \sec ^{2} x}{\tan ^{4} x+1} d x $
$\Rightarrow 2 I=\frac{\pi}{2} \cdot \frac{1}{2} \int\limits_{0}^{\pi / 2} \frac{1}{1+\left(\tan ^{2} x\right)^{2}} d\left(\tan ^{2} x\right) $
$\Rightarrow 2 I=\frac{\pi}{4} \cdot\left[\tan ^{-1} t\right]_{0}^{\infty}=\frac{\pi}{4}\left(\tan ^{-1} \infty-\tan ^{-1} 0\right)$
[where, $t=\tan ^{2} x$ ]
$\Rightarrow I=\frac{\pi^{2}}{16}$