Question

The value of $\underset{0}{\overset{2\pi }{\int }}\frac{x{\sin }^{8}x}{{\sin }^{8}x+{\cos }^{8}x}dx$ is equal to :

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

Answer 1

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

$\text{Let}I=\underset{0}{\overset{2\pi }{\int }}\frac{x{\sin }^{8}x}{{\sin }^{8}x+{\cos }^{8}x}dx...\left(1\right)I=\underset{0}{\overset{2\pi }{\int }}\frac{(2\pi -x){\sin }^8(2\pi -x)}{{\sin }^8(2\pi -x)+{\cos }^8(2\pi -x)}dx=\underset{0}{\overset{2\pi }{\int }}\frac{(2\pi -x){\sin }^{8}x}{{\sin }^{8}x+{\cos }^{8}x}dx...\left(2\right)$

Adding (1) & (2), we get :

$\Rightarrow 2I=2\pi \underset{0}{\overset{2\pi }{\int }}\frac{{\sin }^{8}x}{{\sin }^{8}x+{\cos }^{8}x}dxI=\pi \underset{0}{\overset{2\pi }{\int }}\frac{{\sin }^{8}x}{{\sin }^{8}x+{\cos }^{8}x}dx$
$I=4\pi \underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{{\sin }^{8}x}{{\sin }^{8}x+{\cos }^{8}x}dx...\left(3\right)$
$I=4\pi \underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{{\sin }^8(\frac{\pi }{2}-x)}{{\sin }^8(\frac{\pi }{2}-x)+{\cos }^8(\frac{\pi }{2}-x)}dxI=4\pi \underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{{\cos }^{8}x}{{\sin }^{8}x+{\cos }^{8}x}dx...\left(4\right)$
Adding (3) and (4) , we get :
$I=2\pi \underset{0}{\overset{\frac{\pi }{2}}{\int }}1dx=2\pi \times \frac{\pi }{2}={\pi }^2$