JEE Main 2022 — Mathematics Calculus

Let, $\displaystyle I={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{dx}{(1+{e}^{x})({sm}^{6}x+{\mathrm{cos}}^{6}x)}...(i)$I=∫−2π​2π​​(1+ex)(sm6x+cos6x)dx​...(i)

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

$$I={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{{e}^{x}dx}{(1+{e}^{x})({\mathrm{sin}}^{6}x+{\mathrm{cos}}^{6}x)}...(\mathrm{ii})$$I=∫−2π​2π​​(1+ex)(sin6x+cos6x)exdx​...(ii)

Adding $(i)$(i) and $(\mathrm{ii})$(ii)

$$2I={\int }_{\frac{-\pi }{2}}^{\frac{\pi }{2}}\frac{(1+{e}^{x})dx}{(1+{e}^{x})({\mathrm{sin}}^{6}x+{\mathrm{cos}}^{6}x)}$$2I=∫2−π​2π​​(1+ex)(sin6x+cos6x)(1+ex)dx​

$$I=\frac{1}{2}\times 2{\int }_{0}^{\frac{\pi }{2}}\frac{dx}{({\mathrm{sin}}^{2}x+{\mathrm{cos}}^{2}x)({\mathrm{sin}}^{4}x+{\mathrm{cos}}^{4}x-{\mathrm{sin}}^{2}x{\mathrm{cos}}^{2}x)}$$I=21​×2∫02π​​(sin2x+cos2x)(sin4x+cos4x−sin2xcos2x)dx​

$$=\int _{0}^{\frac{\pi }{2}}\frac{dx}{{({\mathrm{sin}}^{2}x+{\mathrm{cos}}^{2}x)}^{2}-3{\mathrm{sin}}^{2}x{\mathrm{cos}}^{2}x}=\int _{0}^{\frac{\pi }{2}}\frac{dx}{1-\frac{3}{4}{\mathrm{sin}}^{2}2x}$$=∫02π​​(sin2x+cos2x)2−3sin2xcos2xdx​=∫02π​​1−43​sin22xdx​

$$={\int }_{0}^{\frac{\pi }{2}}\frac{4{\mathrm{sec}}^{2}2xdx}{4(1+{\mathrm{tan}}^{2}2x)-3{\mathrm{tan}}^{2}2x}=2{\int }_{0}^{\frac{\pi }{4}}\frac{4{\mathrm{sec}}^{2}2x}{4+{\mathrm{tan}}^{2}2x}dx$$=∫02π​​4(1+tan22x)−3tan22x4sec22xdx​=2∫04π​​4+tan22x4sec22x​dx

Let $\mathrm{tan}2x=t$tan2x=t and $2{\mathrm{sec}}^{2}2xdx=dt$2sec22xdx=dt

$\displaystyle I=4\int _{0}^{\infty }\frac{dt}{{2}^{2}+{t}^{2}}=\frac{4}{2}{[{\mathrm{tan}}^{-1}\frac{t}{2}]}_{0}^{\infty }$I=4∫0∞​22+t2dt​=24​[tan−12t​]0∞​ $=2(\frac{\pi }{2}-0)=\pi$=2(2π​−0)=π