JEE Main 2018 — Mathematics Calculus

$$I= \int _{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{{\mathrm{sin}}^{2}x}{1+{2}^{x}} dx...(i)$$I=∫−2π​2π​​1+2xsin2x​dx...(i)

As $\displaystyle \int _{a}^{b}f(x)dx=\int _{a}^{b}f(a+b-x)dx$∫ab​f(x)dx=∫ab​f(a+b−x)dx

$$I=\int _{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{{\mathrm{sin}}^{2}⁡(-x)}{1+{2}^{-x}}dx$$I=∫−2π​2π​​1+2−xsin2⁡(−x)​dx

$$I= \int _{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{{2}^{x}{\mathrm{sin}}^{2}⁡x}{1+{2}^{x}}dx...(\mathrm{ii})$$I=∫−2π​2π​​1+2x2xsin2⁡x​dx...(ii)

By adding Equations $(i)$(i) and $(\mathrm{ii})$(ii),

$$2I= \int _{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{(1+{2}^{x}){\mathrm{sin}}^{2}⁡x}{(1+{2}^{x})}dx$$2I=∫−2π​2π​​(1+2x)(1+2x)sin2⁡x​dx

$$2I=2\int _{0}^{\frac{\pi }{2}}{\mathrm{sin}}^{2}⁡xdx$$2I=2∫02π​​sin2⁡xdx

$$I=\int _{0}^{\frac{\pi }{2}}{\mathrm{sin}}^{2}⁡xdx...(iii)$$I=∫02π​​sin2⁡xdx...(iii)

$$I=\int _{0}^{\frac{\pi }{2}}{\mathrm{sin}}^{2}⁡(\frac{\pi }{2}-x)dx$$I=∫02π​​sin2⁡(2π​−x)dx

$$I=\int _{0}^{\frac{\pi }{2}}{\mathrm{cos}}^{2}⁡xdx...(iv)$$I=∫02π​​cos2⁡xdx...(iv)

On adding (iii)&(iv), we get

$$2I= \int _{0}^{\frac{\pi }{2}}({\mathrm{sin}}^{2}⁡x+{\mathrm{cos}}^{2}⁡x)dx$$2I=∫02π​​(sin2⁡x+cos2⁡x)dx

$$2I= \int _{0}^{\frac{\pi }{2}}1 dx$$2I=∫02π​​1dx

$$2I= {[x]}_{0}^{\frac{\pi }{2}}$$2I=[x]02π​​

$$2I=\frac{\pi }{2}-0$$2I=2π​−0

$$I=\frac{\pi }{2}\times \frac{1}{2}=\frac{\pi }{4}$$I=2π​×21​=4π​