Question

For $n>0,$ solution of integral ${\int }_0^{2\pi }\frac{x{\sin }^{2n}x}{{\sin }^{2n}x+{\cos }^{2n}x}dx$, is equal to

• A. ${\pi }^2$
• B. $\frac{\pi }{2}$
• C. $2\pi$
• D. $\frac{\pi }{4}$

Answer 1

Answer: (A)

${\pi }^2$

Let $I={\int }_0^{2\pi }\frac{x{\sin }^{2n}x}{{\sin }^{2n}x+{\cos }^{2n}x}dx$ ...(1)

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

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

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

Using (1)
$\Rightarrow I={\int }_0^{2\pi }\frac{\pi {\sin }^{2n}x}{{\sin }^{2n}x+{\cos }^{2n}x}dx$

Now using ${\int }_0^{2a}f\left(x\right)dx={\int }_0^a(f\left(x\right)+f(2a-x))dx$
$I=\pi [{\int }_0^{\pi }\frac{{\sin }^{2n}x}{{\sin }^{2n}x+{\cos }^{2n}x}dx+{\int }_0^{\pi }\frac{{\sin }^{2n}x}{{\sin }^{2n}x+{\cos }^{2n}x}dx]$

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

$\Rightarrow I=4\pi {\int }_0^{\frac{\pi }{2}}\frac{{\sin }^{2n}x}{{\sin }^{2n}x+{\cos }^{2n}x}dx$ ...(2)

$\Rightarrow I=4\pi {\int }_0^{\frac{\pi }{2}}\frac{{\sin }^{2n}(\frac{\pi }{2}-x)}{{\sin }^{2n}(\frac{\pi }{2}-x)+{\cos }^{2n}(\frac{\pi }{2}-x)}dx$

$=4\pi {\int }_0^{\frac{\pi }{2}}\frac{{\cos }^{2n}x}{{\cos }^{2n}x+{\sin }^{2n}x}dx$ ..(3)
Adding (2) and (3), we get

$2I=4\pi {\int }_0^{\frac{\pi }{2}}dx=4\pi .\frac{\pi }{2}\Rightarrow I={\pi }^2$