Let n ≥ 2 be a natural number and 0< < 2 . Then ( ^ n - )^ 1 / n ^ n+1 d is equal…
Mathematics · JEE Main · NTA Exams — Integral Calculus
Let n ≥ 2 be a natural number and \(0<\theta<\frac{\pi}{2}\). Then \(\int \frac{\left(\sin ^{n} \theta-\sin \theta\right)^{1 / n} \cos \theta}{\sin ^{n+1} \theta} d \theta\) is equal to:(where C is a constant of integration)
- \(\frac{n}{n^{2}-1}\left(1+\frac{1}{\sin ^{n-1} \theta}\right)^{\frac{n+1}{n}}+c\)
- \(\frac{n}{n^{2}-1}\left(1-\frac{1}{\sin ^{n+1} \theta}\right)^{\frac{n+1}{n}}+c\)
- \(\frac{n}{n^{2}-1}\left(1-\frac{1}{\sin ^{n-1} \theta}\right)^{\frac{n+1}{n}}+c\)
- \(\frac{n}{n^{2}+1}\left(1-\frac{1}{\sin ^{n-1} \theta}\right)^{\frac{n+1}{n}}+c\)
Answer
(C) n n^ 2 -1 (1- 1 ^ n-1 )^ n+1 n +c
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