JEE Main 2019 (Online) 10th January Morning ShiftQuestion 1
Let $n\ge2$ be a natural number and $0<\theta<\frac\pi2$. Then $\int\frac{\left(\sin^{n}\theta-\sin\theta\right)^{\frac1n}\cos\theta}{\sin^{n+1}\theta}d\theta$ is equal to
Indefinite Integration
Solution
Concept. Take $\sin^n\theta$ out of the $n$-th root; then $u=1-\sin^{1-n}\theta$. Formulas used. (1) $u=1-\dfrac{1}{\sin^{n-1}\theta}\Rightarrow du=(n-1)\dfrac{\cos\theta}{\sin^n\theta}\,d\theta$Step 1 - Rewrite. $$\left(\sin^n\theta-\sin\theta\right)^{1/n}=\sin\theta\left(1-\frac{1}{\sin^{n-1}\theta}\right)^{1/n},$$ so the integrand is $\left(1-\sin^{1-n}\theta\right)^{1/n}\dfrac{\cos\theta}{\sin^n\theta}$. Step 2 - Use (1). $$\frac{1}{n-1}\int u^{1/n}\,du=\frac{1}{n-1}\cdot\frac{n}{n+1}u^{\frac{n+1}{n}}+c.$$ Step 3 - Back to $\theta$. $$\frac{n}{n^2-1}\left(1-\frac{1}{\sin^{n-1}\theta}\right)^{\frac{n+1}{n}}+c.$$ Answer: (C)