JEE Mains · Maths · STD 12 - 7.2 definite integral
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
- A \(50\)
- B \(49\)
- C \(51\)
- D \(25\)
Answer & Solution
Correct Answer
(A) \(50\)
Step-by-step Solution
Detailed explanation
\(\int \limits_0^{\pi / 4} e^{-x} \tan ^{50} x d x\) \(\left[-e^{-y}(\tan x)^{50}\right]_0^{\pi / 4}+\int \limits_0^{\pi / 4} e^{-x}(50)(\tan x)^{49} \sec ^2 x\) \(=-e^{-\pi / 4}+0+50 \int \limits_0^{\pi / 4} e^{-x}(\tan x)^{49}\left(\tan ^2 x+1\right)\)…
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