JEE PYQ: Indefinite Integration Question 29
Question 29 - 2019 (10 Jan Shift 1)
Let $n \ge 2$ be a natural number and $0 < \theta < \frac{\pi}{2}$. Then $\int \frac{(\sin^n\theta + \sin\theta)^{1/n}\cos\theta}{\sin^{n+1}\theta},d\theta$ is equal to (where $C$ is a constant of integration):
(1) $\frac{n}{n^2-1}\left(1-\frac{1}{\sin^{n-1}\theta}\right)^{\frac{n+1}{n}} + C$
(2) $\frac{n}{n^2+1}\left(1-\frac{1}{\sin^{n-1}\theta}\right)^{\frac{n+1}{n}} + C$
(3) $\frac{n}{n^2-1}\left(1+\frac{1}{\sin^{n-1}\theta}\right)^{\frac{n+1}{n}} + C$
(4) $\frac{n}{n^2-1}\left(1-\frac{1}{\sin^{n+1}\theta}\right)^{\frac{n+1}{n}} + C$
Show Answer
Answer: (1)
Solution
Let $\sin\theta = u$, $\cos\theta,d\theta = du$. $I = \int \frac{(u^n + u)^{1/n}}{u^{n+1}},du = \int u^{-n}(1-u^{1-n})^{1/n},du$. Let $1 - u^{1-n} = v$, then $(n-1)u^{-n},du = dv$. $I = \frac{1}{n-1}\int v^{1/n},dv = \frac{n}{(n-1)(n+1)}v^{\frac{n+1}{n}} + C = \frac{n}{n^2-1}\left(1-\frac{1}{\sin^{n-1}\theta}\right)^{\frac{n+1}{n}} + C$.
BETA
