The value of e - 4 + 0 4 e - x tan 50 x ? d x 0 4 e - x ( tan 49 x + tan 51 x ) ? d x is: [2023]

Question

$The value of \frac{e^{- \frac{π}{4}} + \int_{0}^{\frac{π}{4}} e^{- x} \tan^{50} x d x}{\int_{0}^{\frac{π}{4}} e^{- x} (\tan^{49} x + \tan^{51} x) d x} is:$ [2023]

Smart Achievers · Accepted Answer

(4)

$\frac{e^{- π / 4} + \int_{0}^{π / 4} e^{- x} \tan^{50} x d x}{\int_{0}^{π / 4} e^{- x} (\tan^{49} x + \tan^{51} x) d x}$

First, simplify, $\int_{0}^{π / 4} e^{- x} \tan^{50} x d x$

$= {[- e^{- x} {(\tan x)}^{50}]}_{0}^{π / 4} + \int_{0}^{π / 4} e^{- x} (50) \tan^{49} x \cdot \sec^{2} x d x$

$= - e^{- π / 4} + 0 + 50 \int_{0}^{π / 4} e^{- x} {(\tan x)}^{49} (1 + \tan^{2} x) d x$

$= - e^{- π / 4} + 50 \int_{0}^{π / 4} e^{- x} [{(\tan x)}^{51} + {(\tan x)}^{49}] d x$

Now, $\frac{- e^{- π / 4} + \int_{0}^{π / 4} e^{- x} {(\tan x)}^{50} d x}{\int_{0}^{π / 4} e^{- x} (\tan^{49} x + \tan^{51} x) d x}$

$= \frac{50 \int_{0}^{π / 4} e^{- x} (\tan^{51} x + \tan^{49} x) d x}{\int_{0}^{π / 4} e^{- x} (\tan^{49} x + \tan^{51} x) d x} = 50$

$∴$ $\frac{e^{- π / 4} + \int_{0}^{π / 4} e^{- x} \tan^{50} x d x}{\int_{0}^{π / 4} e^{- x} (\tan^{49} x + \tan^{51} x) d x} = 50$

Solution

Q.
$The value of \frac{e^{- \frac{π}{4}} + \int_{0}^{\frac{π}{4}} e^{- x} \tan^{50} x d x}{\int_{0}^{\frac{π}{4}} e^{- x} (\tan^{49} x + \tan^{51} x) d x} is:$ [2023]

1 51

2 49

3 25

4 50

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