JEE Main 2021 — Mathematics Calculus

Given $\displaystyle I={\int }_{-\pi /2}^{\pi /2}\frac{{\mathrm{cos}}^{2}x}{1+{3}^{x}}dx...(i)$I=∫−π/2π/2​1+3xcos2x​dx...(i)

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

$$I={\int }_{-\pi /2}^{\pi /2}\frac{{\mathrm{cos}}^{2}x}{1+{3}^{-x}}dx={\int }_{-\pi /2}^{\pi /2}\frac{{3}^{x}{\mathrm{cos}}^{2}x}{1+{3}^{x}}dx...(ii)$$I=∫−π/2π/2​1+3−xcos2x​dx=∫−π/2π/2​1+3x3xcos2x​dx...(ii)

Adding (i)&(ii) we get,

$$2I={\int }_{-\pi /2}^{\pi /2}\frac{(1+{3}^{x}){\mathrm{cos}}^{2}x}{1+{3}^{x}}dx$$2I=∫−π/2π/2​1+3x(1+3x)cos2x​dx

$$2I={\int }_{-\pi /2}^{\pi /2}{\mathrm{cos}}^{2}xdx$$2I=∫−π/2π/2​cos2xdx

$$2I=2{\int }_{0}^{\pi /2}{\mathrm{cos}}^{2}xdx$$2I=2∫0π/2​cos2xdx

$$I={\int }_{0}^{\pi /2}{\mathrm{cos}}^{2}xdx$$I=∫0π/2​cos2xdx

$$I=\frac{1}{2}{\int }_{0}^{\pi /2}(1+\mathrm{cos}2x)dx=\frac{1}{2}{[x+\frac{\mathrm{sin}2x}{2}]}_{0}^{\pi /2}$$I=21​∫0π/2​(1+cos2x)dx=21​[x+2sin2x​]0π/2​

$$I=\frac{\pi }{4}$$I=4π​