The value of e^ - 4 + _0^ 4 e^ -x ^ 50 x d x _0^ 4 e^ -x ( ^ 49 x… - Vidyadip

Question

The value of $\frac{e^{-\frac{\pi}{4}}+\int \limits_0^{\frac{\pi}{4}} e^{-x} \tan ^{50} x d x}{\int \limits_0^{\frac{\pi}{4}} e^{-x}\left(\tan ^{49} x+\tan ^{51} x\right) d x}$ is

✓

Answer

a
$\int \limits_0^{\pi / 4} e^{-x} \tan ^{50} x d x$

$\left[-e^{-y}(\tan x)^{50}\right]_0^{\pi / 4}+\int \limits_0^{\pi / 4} e^{-x}(50)(\tan x)^{49} \sec ^2 x$

$=-e^{-\pi / 4}+0+50 \int \limits_0^{\pi / 4} e^{-x}(\tan x)^{49}\left(\tan ^2 x+1\right)$

$=-e^{-\pi / 4}+50\left(\int \limits_0^{\pi / 4} e^{-x}(\tan x)^{51}+(\tan x)^{49}\right) d x$

Now,$\frac{-e^{-\pi / 4}+\int \limits_0^{\pi / 4} e^{-x}(\tan x)^{50} d x}{\int \limits_0^{\pi / 4} e^{-x}\left(\tan ^{49} x+\tan ^{51} x\right) d x}$

$\frac{50 \int \limits_0^{\pi / 4} e^{-x}\left((\tan x)^{51}+(\tan x)^{49}\right) d x}{\int \limits_0^{\pi / 4} e^{-x}\left(\tan ^{49} x+\tan ^{51} x\right) d x}=50$

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