JEE Main 2021 — Mathematics Calculus

$$I={\int }_{a}^{b}f(x)dx$$I=∫ab​f(x)dx

$$f(a+b-x)=f(x)$$f(a+b−x)=f(x)

$$I={\int }_{\frac{-\pi }{2}}^{\frac{\pi }{2}}\frac{1+{\mathrm{sin}}^{2}x}{1+{\pi }^{\mathrm{sin}x}}dx\ldots (i)$$I=∫2−π​2π​​1+πsinx1+sin2x​dx…(i)

$$I={\int }_{\frac{-\pi }{2}}^{\frac{\pi }{2}}\frac{1+{\mathrm{sin}}^{2}x}{1+{\pi }^{-\mathrm{sin}x}}dx...(\mathrm{ii})$$I=∫2−π​2π​​1+π−sinx1+sin2x​dx...(ii)

Add equation $(i)$(i) and $(\mathrm{ii})$(ii)

$$2I={\int }_{\frac{-\pi }{2}}^{\frac{\pi }{2}}(1+{\mathrm{sin}}^{2}x)dx$$2I=∫2−π​2π​​(1+sin2x)dx

$$2I=\pi +\frac{1}{2}{\int }_{\frac{-\pi }{2}}^{\frac{\pi }{2}}(1-\mathrm{cos}2x)dx$$2I=π+21​∫2−π​2π​​(1−cos2x)dx

$$2I=\pi +\frac{1}{2}{[x-\frac{\mathrm{sin}2x}{2}]}_{\frac{-\pi }{2}}^{\frac{\pi }{2}}$$2I=π+21​[x−2sin2x​]2−π​2π​​

$$2I=\pi +\frac{1}{2}[(\frac{\pi }{2}-0)-(\frac{-\pi }{2}-0)]$$2I=π+21​[(2π​−0)−(2−π​−0)]

$$2I=\pi +\frac{1}{2}\pi$$2I=π+21​π

$$2I=\frac{3\pi }{2}$$2I=23π​

$$\Rightarrow I=\frac{3\pi }{4}$$⇒I=43π​