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

Answer 1

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

Let $I=\underset{0}{\overset{2\pi }{\int }}\frac{x{\sin }^{8}x}{{\sin }^{8}x+{\cos }^{8}x}dx\cdots \left(i\right)$
Using property
$\underset{a}{\overset{b}{\int }}f\left(x\right)dx=\underset{a}{\overset{b}{\int }}f(a+b-x)dx$
$\Rightarrow I=\underset{0}{\overset{2\pi }{\int }}\frac{(2\pi -x){\sin }^{8}x}{{\sin }^{8}x+{\cos }^{8}x}dx\cdots \left(ii\right)$
Adding $\left(i\right)$ and $\left(ii\right)$, we get
$2I=2\pi \underset{0}{\overset{2\pi }{\int }}\frac{{\sin }^{8}x}{{\sin }^{8}x+{\cos }^{8}x}dx$
$\Rightarrow I=4\pi \underset{0}{\overset{\pi /2}{\int }}\frac{{\sin }^{8}x}{{\sin }^{8}x+{\cos }^{8}x}dx$
$\{\because \underset{0}{\overset{2a}{\int }}f\left(x\right)dx=2\underset{0}{\overset{a}{\int }}f\left(x\right)dx,\text{when}f\left(x\right)=f(2a-x)\}$
$\Rightarrow I=4\pi \cdot \frac{\pi }{4}={\pi }^2$
$(\because \underset{0}{\overset{\pi /2}{\int }}\frac{{\sin }^{n}x}{{\cos }^{n}x+{\sin }^{n}x}dx=\frac{\pi }{4})$