Question

${\int }_0^{\pi /2}\frac{x\sin x\cdot \cos x}{{\cos }^{4}x+{\sin }^{4}x}dx$ is equal to

• A. $\frac{{\pi }^2}{8}$
• B. $\frac{{\pi }^2}{16}$
• C. $1$
• D. $0$

Answer 1

Answer: (B)

Let $I={\int }_0^{\pi /2}\frac{x\sin x\cos x}{{\cos }^{4}x+{\sin }^{4}x}dx$ .. (i)
$\Rightarrow$ $I={\int }_0^{\pi /2}\frac{\left(\frac{\pi }{2}-x\right)\sin x\cos x}{{\sin }^{4}x+{\cos }^{4}x}dx$ ... (ii)
On adding Eqs. (i) and (ii), we get
$2I=\frac{\pi }{2}{\int }_0^{\pi /2}\frac{\sin x\cos x}{{\cos }^{4}x+{\sin }^{4}x}dx$
$\Rightarrow$ $I=\frac{\pi }{4}{\int }_0^{\pi /2}\frac{\tan x{\sec }^{2}x}{1+{\tan }^{4}x}$
(dividing $Nr$ and $Dr$ by ${\cos }^{4}x)$
Put, ${\tan }^{2}x=t\Rightarrow 2\tan x{\sec }^{2}xdx=dt$
$\therefore$ $I=\frac{\pi }{4}{\int }_0^{\infty }\frac{1}{2(1+t^2)}dt=\frac{\pi }{8}[{\tan }^{-1}t{]}_0^{\infty }$
$=\frac{\pi }{8}\times \frac{\pi }{2}=\frac{{\pi }^2}{16}$