Question

The value of ${\int }_0^{\pi /2}\frac{x\sin x\cos x}{{\sin }^{4}x+{\cos }^{4}x}dx$ is

• A. ${\pi }^2/8$
• B. ${\pi }^2/16$
• C. $3{\pi }^2/4$
• D. ${\pi }^2/2$

Answer 1

The correct option is B ${\pi }^2/16$

${\int }_0^{\pi /2}\frac{x\sin x\cos x}{{\sin }^{4}x+{\cos }^{4}x}dx$
$={\int }_0^{\pi /2}\frac{(\pi /2-x)\sin x\cos x}{{\sin }^{4}x+{\cos }^{4}x}dx$
$=\frac{\pi }{2}{\int }_0^{\pi /2}\frac{\sin x\cos x}{{\sin }^{4}x+{\cos }^{4}x}dx-{\int }_0^{\pi /2}\frac{x\sin x\cos x}{{\sin }^{4}x+{\cos }^{4}x}dx$
$\Rightarrow 2I=\frac{\pi }{2}{\int }_0^{\pi /2}\frac{\sin x\cos x}{{\sin }^{4}x+{\cos }^{4}x}dx$
$\Rightarrow I=\frac{\pi /}{4}{\int }_0^{\pi /2}\frac{1}{2}\cdot \frac{\sin 2x}{1-\frac{1}{2}{\sin }^{2}2x}=\frac{\pi }{4}{\int }_0^{\pi /2}\frac{\sin 2x}{1+{\cos }^{2}2x}dx$
$=\frac{-\pi }{8}{\int }_1^{-1}\frac{dt}{1+t^2}=\frac{-\pi }{8}\left[{\tan }^{-1}\right(-1)-{\tan }^{-1}(1\left)\right]=\frac{{\pi }^2}{16}$