Question

Evaluate ${\int }_0^{2\pi }\frac{x{\sin }^{2n}x}{{\sin }^{2n}x+{\cos }^{2n}x}dx,$ for $n>0$,

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

Answer 1

The correct option is C ${\pi }^2$

$I={\int }_0^{2\pi }\frac{x{\sin }^{2n}x}{{\sin }^{2n}x+{\cos }^{2n}x}dx$

$\Rightarrow I={\int }_0^{2\pi }\frac{(2\pi -x){\sin }^{2n}x}{{\sin }^{2n}x+{\cos }^{2n}x}dx$

.

Hence $2I={\int }_0^{2\pi }\frac{2\pi {\sin }^{2n}x}{{\sin }^{2n}x+{\cos }^{2n}x}dx$

$\Rightarrow I=\pi {\int }_0^{2\pi }\frac{{\sin }^{2n}x}{{\sin }^{2n}x+{\cos }^{2n}x}dx$

$\Rightarrow I=\pi \times 2{\int }_0^{\pi }\frac{{\sin }^{2n}x}{{\sin }^{2n}x+{\cos }^{2n}x}dx$ $(\because$ the given function is periodic with period $\pi$)

Substitute $t=\frac{\pi }{2}-x$

$\Rightarrow dt=-dx$

$\Rightarrow I=2\pi {\int }_{\pi /2}^{-\pi /2}\frac{{\cos }^{2n}t}{{\sin }^{2n}t+{\cos }^{2n}t}(-dt)$

$\Rightarrow I=2\pi {\int }_{-\pi /2}^{\pi /2}\frac{{\cos }^{2n}x}{{\sin }^{2n}x+{\cos }^{2n}x}dx$

$\Rightarrow I=4\pi {\int }_0^{\pi /2}\frac{{\cos }^{2n}x}{{\sin }^{2n}x+{\cos }^{2n}x}dx$......(1) ($\because f\left(x\right)$ is an even function)

$\Rightarrow I=4\pi {\int }_0^{\pi /2}\frac{{\sin }^{2n}x}{{\sin }^{2n}x+{\cos }^{2n}x}dx$.....(2) (replacing $f\left(x\right)$ by $f(\frac{\pi }{2}-x)$)

Adding (1) and (2), we get

$2I=4\pi {\int }_0^{\pi /2}1.dx$

$\Rightarrow I=2\pi \times \frac{\pi }{2}={\pi }^2$