JEE Main 2021 — Mathematics Calculus

$$I={\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}\frac{dx}{(1+{e}^{x\mathrm{cos}x})({\mathrm{sin}}^{4}x+{\mathrm{cos}}^{4}x)}\ldots (1)$$I=∫−4π​4π​​(1+excosx)(sin4x+cos4x)dx​…(1)

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

$$I={\int }_{-\pi /4}^{\pi /4}\frac{dx}{(1+{e}^{-x\mathrm{cos}x})({\mathrm{sin}}^{4}x+{\mathrm{cos}}^{4}x)}...(2)$$I=∫−π/4π/4​(1+e−xcosx)(sin4x+cos4x)dx​...(2)

Add $(1)$(1) and $(2)$(2)

$$2I={\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}\frac{dx}{{\mathrm{sin}}^{4}x+{\mathrm{cos}}^{4}x}$$2I=∫−4π​4π​​sin4x+cos4xdx​

$$\Rightarrow 2I=2{\int }_{0}^{\frac{\pi }{4}}\frac{dx}{{\mathrm{sin}}^{4}x+{\mathrm{cos}}^{4}x}$$⇒2I=2∫04π​​sin4x+cos4xdx​

$$\Rightarrow I={\int }_{0}^{\frac{\pi }{4}}\frac{(1+{\mathrm{tan}}^{2}x){\mathrm{sec}}^{2}x}{{\mathrm{tan}}^{4}x+1}dx$$⇒I=∫04π​​tan4x+1(1+tan2x)sec2x​dx

$$\Rightarrow I={\int }_{0}^{\frac{\pi }{4}}\frac{(1+\frac{1}{{\mathrm{tan}}^{2}x}){\mathrm{sec}}^{2}x}{{(\mathrm{tan}x-\frac{1}{\mathrm{tan}x})}^{2}+2}dx$$⇒I=∫04π​​(tanx−tanx1​)2+2(1+tan2x1​)sec2x​dx

Put $\mathrm{tan}x-\frac{1}{\mathrm{tan}x}=t$tanx−tanx1​=t

$$\Rightarrow (1+\frac{1}{{\mathrm{tan}}^{2}x}){\mathrm{sec}}^{2}xdx=dt$$⇒(1+tan2x1​)sec2xdx=dt

So, $\displaystyle I={\int }_{-\infty }^{0}\frac{dt}{{t}^{2}+2}={[\frac{1}{\sqrt{2}}{\mathrm{tan}}^{-1}(\frac{t}{\sqrt{2}})]}_{-\infty }^{0}$I=∫−∞0​t2+2dt​=[2​1​tan−1(2​t​)]−∞0​

$$\Rightarrow I=0-\frac{1}{\sqrt{2}}(-\frac{\pi }{2})=\frac{\pi }{2\sqrt{2}}$$⇒I=0−2​1​(−2π​)=22​π​