The value of frac{e^{-pi/4} + int_0^{pi/4} e^ | vedclass

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(A) Let $I = \int_0^{\pi/4} e^{-x} 	an^{50} x \, dx$. Using integration by parts with $u = 	an^{50} x$ and $dv = e^{-x} \, dx$,we have $du = 50 	an

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$50$

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(A) Let $I = \int_0^{\pi/4} e^{-x} 	an^{50} x \, dx$. Using integration by parts with $u = 	an^{50} x$ and $dv = e^{-x} \, dx$,we have $du = 50 	an^{49} x \sec^2 x \, dx$ and $v = -e^{-x}$.$I = [-e^{-x} 	an^{50} x]_0^{\pi/4} + \int_0^{\pi/4} e^{-x} (50 	an^{49} x \sec^2 x) \, dx$$I = -e^{-\pi/4} (1)^{50} + 0 + 50 \int_0^{\pi/4} e^{-x} 	an^{49} x (1 + 	an^2 x) \, dx$$I = -e^{-\pi/4} + 50 \int_0^{\pi/4} e^{-x} (	an^{49} x + 	an^{51} x) \, dx$Rearranging the terms,we get:$e^{-\pi/4} + I = 50 \int_0^{\pi/4} e^{-x} (	an^{49} x + 	an^{51} x) \, dx$Therefore,the given expression is:$\frac{e^{-\pi/4} + I}{\int_0^{\pi/4} e^{-x} (	an^{49} x + 	an^{51} x) \, dx} = \frac{50 \int_0^{\pi/4} e^{-x} (	an^{49} x + 	an^{51} x) \, dx}{\int_0^{\pi/4} e^{-x} (	an^{49} x + 	an^{51} x) \, dx} = 50$

The value of $\frac{e^{-\pi/4} + \int_0^{\pi/4} e^{-x} \tan^{50} x \, dx}{\int_0^{\pi/4} e^{-x} (\tan^{49} x + \tan^{51} x) \, dx}$ is

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