JEE Main 2020 — Mathematics Calculus

Given, $\displaystyle I={\int }_{0}^{2\pi }\frac{x{\mathrm{sin}}^{8}⁡x}{{\mathrm{sin}}^{8}⁡x+{\mathrm{cos}}^{8}⁡x}dx...(i)$I=∫02π​sin8⁡x+cos8⁡xxsin8⁡x​dx...(i)

Use the property $\displaystyle {\int }_{a}^{b}f(x)dx={\int }_{a}^{b}f(a+b-x)dx$∫ab​f(x)dx=∫ab​f(a+b−x)dx we get,

$$I={\int }_{0}^{2\pi }\frac{(2\pi -x){\mathrm{sin}}^{8}⁡(2\pi -x)}{{\mathrm{sin}}^{8}⁡(2\pi -x)+{\mathrm{cos}}^{8}⁡(2\pi -x)}dx={\int }_{0}^{2\pi }\frac{(2\pi -x){\mathrm{sin}}^{8}⁡x}{{\mathrm{sin}}^{8}⁡x+{\mathrm{cos}}^{8}⁡x}dx...(ii)$$I=∫02π​sin8⁡(2π−x)+cos8⁡(2π−x)(2π−x)sin8⁡(2π−x)​dx=∫02π​sin8⁡x+cos8⁡x(2π−x)sin8⁡x​dx...(ii)

Adding equation (i)&(ii)we get,

$$2I={\int }_{0}^{2\pi }\frac{2\pi {\mathrm{sin}}^{8}⁡x}{{\mathrm{sin}}^{8}⁡x+{\mathrm{cos}}^{8}⁡x}dx$$2I=∫02π​sin8⁡x+cos8⁡x2πsin8⁡x​dx

Use the property $\displaystyle {\int }_{0}^{2a}f(x)dx=2{\int }_{0}^{a}f(2a-x)dx$∫02a​f(x)dx=2∫0a​f(2a−x)dx we get,

$$I=2{\int }_{0}^{\pi }\frac{\pi {\mathrm{sin}}^{8}⁡x}{{\mathrm{sin}}^{8}⁡x+{\mathrm{cos}}^{8}⁡x}dx$$I=2∫0π​sin8⁡x+cos8⁡xπsin8⁡x​dx

Use the property $\displaystyle {\int }_{0}^{2a}f(x)dx=2{\int }_{0}^{a}f(2a-x)dx$∫02a​f(x)dx=2∫0a​f(2a−x)dxwe get,

$$I=4{\int }_{0}^{\pi /2}\frac{\pi {\mathrm{sin}}^{8}⁡x}{{\mathrm{sin}}^{8}⁡x+{\mathrm{cos}}^{8}⁡x}dx...(iii)$$I=4∫0π/2​sin8⁡x+cos8⁡xπsin8⁡x​dx...(iii)

Use the property $\displaystyle {\int }_{a}^{b}f(x)dx={\int }_{a}^{b}f(a+b-x)dx$∫ab​f(x)dx=∫ab​f(a+b−x)dx we get,

$$I=4{\int }_{0}^{\pi /2}\frac{\pi {\mathrm{cos}}^{8}⁡x}{{\mathrm{sin}}^{8}⁡x+{\mathrm{cos}}^{8}⁡x}dx...(iv)(\because \mathrm{sin}(\frac{\pi }{2}-x)=\mathrm{cos}x)$$I=4∫0π/2​sin8⁡x+cos8⁡xπcos8⁡x​dx...(iv)(∵sin(2π​−x)=cosx)

Adding the equation (iii)&(iv)we get,

$$2I=4\pi {\int }_{0}^{\pi /2}1dx$$2I=4π∫0π/2​1dx

$$\Rightarrow I=2\pi {[x]}_{0}^{\pi /2}={\pi }^{2}$$⇒I=2π[x]0π/2​=π2