JEE Main 2022 — Mathematics Calculus

Given $\displaystyle I=\int _{0}^{\pi }\frac{\mathrm{sin}x\cdot {e}^{\mathrm{cos}x}}{(1+{\mathrm{cos}}^{2}x)({e}^{\mathrm{cos}x}+{e}^{-\mathrm{cos}x})}dx...(i)$I=∫0π​(1+cos2x)(ecosx+e−cosx)sinx⋅ecosx​dx...(i)

Using property of integral $\displaystyle \int f_{a}^{b}(a+b-x)=\int _{a}^{b}f(x)$∫fab​(a+b−x)=∫ab​f(x)

we get $\displaystyle I=\int _{0}^{\pi }\frac{\mathrm{sin}x\cdot {e}^{-\mathrm{cos}x}}{(1+{\mathrm{cos}}^{2}x)({e}^{-\mathrm{cos}x}+{e}^{\mathrm{cos}x})}dx...(\mathrm{ii})$I=∫0π​(1+cos2x)(e−cosx+ecosx)sinx⋅e−cosx​dx...(ii)

Adding equation $(i)$(i) & $(\mathrm{ii})$(ii) we get

$$2I=\int _{0}^{\pi }\frac{\mathrm{sin}x}{(1+{\mathrm{cos}}^{2}x)}\cdot \frac{({e}^{\mathrm{cos}x}+{e}^{-\mathrm{cos}x})}{{e}^{\mathrm{cos}x}+{e}^{-\mathrm{cos}x}}dx$$2I=∫0π​(1+cos2x)sinx​⋅ecosx+e−cosx(ecosx+e−cosx)​dx

$\displaystyle 2I=\int _{0}^{\pi }\frac{\mathrm{sin}x}{1+{\mathrm{cos}}^{2}x}dx$2I=∫0π​1+cos2xsinx​dx $\displaystyle \Rightarrow 2I=2\int _{0}^{\frac{\pi }{2}}\frac{\mathrm{sin}x}{1+{\mathrm{cos}}^{2}x}dx$⇒2I=2∫02π​​1+cos2xsinx​dx

Let $\mathrm{cos}x=t\Rightarrow -\mathrm{sin}xdx=dt$cosx=t⇒−sinxdx=dt

$\displaystyle I=-\int _{1}^{0}\frac{dt}{1+{t}^{2}}\Rightarrow I=-{[{\mathrm{tan}}^{-1}t]}_{1}^{0}=-[0-\frac{\pi }{4}]$I=−∫10​1+t2dt​⇒I=−[tan−1t]10​=−[0−4π​] $=\frac{\pi }{4}$=4π​