WBJEE · Maths · Definite Integration
The value of the integral \(\int_{0}^{\pi / 2} \frac{1}{1+(\tan x)^{-101}} d x\) is equal to
- A 1
- B \(\frac{\pi}{6}\)
- C \(\frac{\pi}{8}\)
- D \(\frac{\pi}{4}\)
Answer & Solution
Correct Answer
(D) \(\frac{\pi}{4}\)
Step-by-step Solution
Detailed explanation
Let \(I=\int_{0}^{\pi / 2} \frac{1}{1+(\tan x)^{-101}} d x\) \(=\int_{0}^{\pi / 2} \frac{d x}{1+\left\{\tan \left(\frac{\pi}{2}-x\right)\right\}^{101}}\) \(=\int_{0}^{\pi / 2} \frac{d x}{1+(\cot x)^{101}}\) \(=\int_{0}^{\pi / 2} \frac{\tan x^{10 t}}{\tan x^{101}+1} d x\)…
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