JEE Main 2023 — Mathematics Calculus

Let,

$$I=\frac{{e}^{-\frac{\pi }{4}}+{\int }_{0}^{\frac{\pi }{4}}{e}^{-x}\mathrm{tan}{}^{50}xdx}{{\int }_{0}^{\frac{\pi }{4}}{e}^{-x}({\mathrm{tan}}^{49}x+{\mathrm{tan}}^{51}x)dx}$$I=∫04π​​e−x(tan49x+tan51x)dxe−4π​+∫04π​​e−xtan50xdx​

Now let $\displaystyle {I}_{1}={\int }_{0}^{\frac{\pi }{4}}{e}^{-x}({\mathrm{tan}}^{49}x+{\mathrm{tan}}^{51}x)dx$I1​=∫04π​​e−x(tan49x+tan51x)dx

Now solving, $\displaystyle {I}_{2}={\int }_{0}^{\frac{\pi }{4}}{e}^{-x}{\mathrm{tan}}^{50}xdx$I2​=∫04π​​e−xtan50xdx using byparts we get,

$${I}_{2}={\int }_{0}^{\frac{\pi }{4}}{e}^{-x}{\mathrm{tan}}^{50}xdx$$I2​=∫04π​​e−xtan50xdx

$$\Rightarrow {I}_{2}={[-{e}^{-x}{\mathrm{tan}}^{50}x]}_{0}^{\frac{\pi }{4}}-{\int }_{0}^{\frac{\pi }{4}}50{\mathrm{tan}}^{49}x{\mathrm{sec}}^{2}x(-{e}^{-x})dx$$⇒I2​=[−e−xtan50x]04π​​−∫04π​​50tan49xsec2x(−e−x)dx

$$\Rightarrow {I}_{2}=-{e}^{-\frac{\pi }{4}}+{\int }_{0}^{\frac{\pi }{4}}50{\mathrm{tan}}^{49}x(1+{\mathrm{tan}}^{2}x){e}^{-x}dx$$⇒I2​=−e−4π​+∫04π​​50tan49x(1+tan2x)e−xdx

$$\Rightarrow {I}_{2}=-{e}^{-\frac{\pi }{4}}+50{\int }_{0}^{\frac{\pi }{4}}({\mathrm{tan}}^{49}x+{\mathrm{tan}}^{51}x){e}^{-x}dx$$⇒I2​=−e−4π​+50∫04π​​(tan49x+tan51x)e−xdx

$$\Rightarrow {I}_{2}=-{e}^{-\frac{\pi }{4}}+50{I}_{1}$$⇒I2​=−e−4π​+50I1​

$$\Rightarrow {\int }_{0}^{\frac{\pi }{4}}{e}^{-x}{\mathrm{tan}}^{50}xdx+{e}^{-\frac{\pi }{4}}=50{I}_{1}$$⇒∫04π​​e−xtan50xdx+e−4π​=50I1​

Now putting the given integral $I$I we get,

$$I=\frac{{e}^{-\frac{\pi }{4}}+{\int }_{0}^{\frac{\pi }{4}}{e}^{-x}\mathrm{tan}{}^{50}xdx}{{\int }_{0}^{\frac{\pi }{4}}{e}^{-x}({\mathrm{tan}}^{49}x+{\mathrm{tan}}^{51}x)dx}$$I=∫04π​​e−x(tan49x+tan51x)dxe−4π​+∫04π​​e−xtan50xdx​

$$\Rightarrow I=\frac{50{I}_{1}}{{I}_{1}}=50$$⇒I=I1​50I1​​=50