WBJEE · Maths · Definite Integration
The value of \(I=\int_{0}^{\frac{\pi}{4}}\left(\tan ^{n+1} x\right) d x\)
\(+\frac{1}{2} \int_{0}^{\frac{\pi}{2}} \tan ^{n+1}\left(\frac{x}{2}\right) d x\) is
- A \(\frac{1}{n}\)
- B \(\frac{n+2}{2 n+1}\)
- C \(\frac{2 n-1}{n}\)
- D \(\frac{2 n-3}{3 n-2}\)
Answer & Solution
Correct Answer
(A) \(\frac{1}{n}\)
Step-by-step Solution
Detailed explanation
.Given. \(I=\int_{0}^{\pi / 4}\left(\tan ^{n+1} x\right) d x+\frac{1}{2} \int_{0}^{\pi / 2} \tan ^{n-1}\left(\frac{x}{2}\right) d x\) In second integral, putt \(=\frac{x}{2} \Rightarrow d x=2 d t\) \(\Rightarrow \quad\) Also, when \(x=0\) then \(t=0\) When \(x=\pi / 2,\) then…
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