JEE Main 2020 — Definite Integration Question with Solution

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

Use the property ${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f\left(a+b-x\right)dx$ we get, 

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

Adding equation $\left(i\right) \& \left(ii\right)$we get, 

$2I={\int }_0^{2\pi }\frac{2\pi {\sin }^{8}x}{{\sin }^{8}x+{\cos }^{8}x}dx$

Use the property ${\int }_0^{2a}f\left(x\right)dx=2{\int }_0^{a}f\left(2a-x\right)dx$ we get, 

$I=2{\int }_0^{\pi }\frac{\pi {\sin }^{8}x}{{\sin }^{8}x+{\cos }^{8}x}dx$

Use the property ${\int }_0^{2a}f\left(x\right)dx=2{\int }_0^{a}f\left(2a-x\right)dx$we get, 

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

Use the property ${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f\left(a+b-x\right)dx$ we get, 

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

Adding the equation $\left(iii\right) \& \left(iv\right)$we get, 

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

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