Question

The value of ${\int }_0^{2\mathrm{\pi }}\frac{x{\sin }^{8}x}{{\sin }^{8}x+{\cos }^{8}x}dx$ equal to?

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

Answer 1

The correct option is D

${\mathrm{\pi }}^2$

Explanation for the correct answer:

Step 1: Compute the required value.

Let $I=$${\int }_0^{2\mathrm{\pi }}\frac{x{\sin }^{8}x}{{\sin }^{8}x+{\cos }^{8}x}dx$ $...\left(i\right)$

$\Rightarrow$ $I=$${\int }_0^{2\mathrm{\pi }}\frac{\left(2\mathrm{\pi }-x\right){\sin }^8\left(2\mathrm{\pi }-x\right)}{{\sin }^8\left(2\mathrm{\pi }-x\right)+{\cos }^8\left(2\mathrm{\pi }-x\right)}dx$ $\left[\because {\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f\left(a-x\right)dx\right]$

$\Rightarrow$ $I=$ ${\int }_0^{2\mathrm{\pi }}\frac{\left(2\mathrm{\pi }-x\right){\sin }^8\left(x\right)}{{\sin }^8\left(x\right)+{\cos }^8\left(x\right)}dx$ $...\left(ii\right)$

Step 2: Add the equation $\left(i\right),\left(ii\right)$

$\Rightarrow$ $2I=$${\int }_0^{2\mathrm{\pi }}\frac{\left(2\mathrm{\pi }\right){\sin }^8\left(x\right)}{{\sin }^8\left(x\right)+{\cos }^8\left(x\right)}dx$

$\Rightarrow$ $I=$$\mathrm{\pi }{\int }_0^{2\mathrm{\pi }}\frac{{\sin }^8\left(x\right)}{{\sin }^8\left(x\right)+{\cos }^8\left(x\right)}\mathrm{dx}$

$\Rightarrow$ $I=$$2\mathrm{\pi }{\int }_0^{\mathrm{\pi }}\frac{{\sin }^8\left(x\right)}{{\sin }^8\left(x\right)+{\cos }^8\left(x\right)}\mathrm{dx}$ $\left[\because {\int }_0^{2a}f\left(x\right)dx=2{\int }_0^{a}f\left(x\right)dxiff\left(a-x\right)=f\left(x\right)\right]$

$\Rightarrow$ $I=$$4\mathrm{\pi }{\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{{\sin }^8\left(x\right)}{{\sin }^8\left(x\right)+{\cos }^8\left(x\right)}\mathrm{dx}$$...\left(iii\right)$ $\left[\because {\int }_0^{2a}f\left(x\right)dx=2{\int }_0^{a}f\left(x\right)dxiff\left(a-x\right)=f\left(x\right)\right]$

$\Rightarrow$ $I=$$4\mathrm{\pi }{\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{{\sin }^8\left({\displaystyle \frac{\mathrm{\pi }}{2}}-x\right)}{{\sin }^8\left({\displaystyle \frac{\mathrm{\pi }}{2}}-x\right)+{\cos }^8\left({\displaystyle \frac{\mathrm{\pi }}{2}}-x\right)}\mathrm{dx}$ $\left[\because {\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f\left(a-x\right)dx\right]$

$\Rightarrow$ $I=$$4\mathrm{\pi }{\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{{\cos }^8\left(x\right)}{{\cos }^8\left(x\right)+{\sin }^8\left(x\right)}\mathrm{dx}$ $...\left(iv\right)$ $\left[\because \sin \left(\frac{\mathrm{\pi }}{2}-x\right)=\cos x,\cos \left(\frac{\mathrm{\pi }}{2}-x\right)=\sin x\right]$

Step 3: Add the equation $\left(iii\right),\left(iv\right)$

$\Rightarrow$ $2I=$$4\mathrm{\pi }{\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{{\cos }^8\left(x\right)+{\sin }^8\left(x\right)}{{\cos }^8\left(x\right)+{\sin }^8\left(x\right)}\mathrm{dx}$

$\Rightarrow$ $I=$$2\mathrm{\pi }{\int }_0^{\frac{\mathrm{\pi }}{2}}1\mathrm{dx}$

$\Rightarrow$ $I=$$2\mathrm{\pi }{\left[\mathrm{x}\right]}_0^{\frac{\mathrm{\pi }}{2}}$

$\Rightarrow$ $I=$$2\mathrm{\pi }\left(\frac{\mathrm{\pi }}{2}-0\right)$

$\Rightarrow$ $I=$${\mathrm{\pi }}^2$

Hence, the value of ${\int }_0^{2\mathrm{\pi }}\frac{x{\sin }^{8}x}{{\sin }^{8}x+{\cos }^{8}x}dx$ is ${\mathrm{\pi }}^2$

Hence, option (D) is the correct answer.