# let n>2

Question:

Let $\mathrm{n} \geq 2$ be a natural number and $0<\theta<\frac{\pi}{2}$ Then $\int \frac{\left(\sin ^{n} \theta+\sin \theta\right)^{\frac{1}{n}} \cos \theta}{\sin ^{n+1} \theta} d \theta$ is equal to:

(where $\mathrm{C}$ is a constant of integration)

1. (1) $\frac{\mathrm{n}}{\mathrm{n}^{2}-1}\left(1-\frac{1}{\sin ^{\mathrm{n}-1} \theta}\right)^{\frac{\mathrm{n}+1}{\mathrm{n}}}+\mathrm{C}$

2. (2) $\frac{\mathrm{n}}{\mathrm{n}^{2}+1}\left(1-\frac{1}{\sin ^{\mathrm{n}-1} \theta}\right)^{\frac{\mathrm{n}+1}{\mathrm{n}}}+\mathrm{C}$

3. (3) $\frac{\mathrm{n}}{\mathrm{n}^{2}-1}\left(1+\frac{1}{\sin ^{\mathrm{n}-1} \theta}\right)^{\frac{\mathrm{n}+1}{\mathrm{n}}}+\mathrm{C}$

4. (4) $\frac{\mathrm{n}}{\mathrm{n}^{2}-1}\left(1-\frac{1}{\sin ^{\mathrm{n}+1} \theta}\right)^{\frac{\mathrm{n}+1}{\mathrm{n}}}+\mathrm{C}$

Correct Option: 1

Solution:

Let, $I=\int \frac{\left(\sin ^{n} \theta-\sin \theta\right)^{\frac{1}{n}} \cos \theta}{\sin ^{n+1} \theta} d \theta$

Let $\sin \theta=u$

$\Rightarrow \cos \theta d \theta=d u$

$\therefore \quad I=\int \frac{\left(u^{n}-u\right)^{\frac{1}{n}}}{u^{n+1}} d u$

$=\int \frac{\left(1-\frac{1}{u^{n-1}}\right)^{\frac{1}{n}}}{u^{n}} d u=\int u^{-n}\left(1-u^{1-n}\right)^{\frac{1}{n}} d u$

Let $1-u^{1-n}=v$

$\Rightarrow-(1-n) u^{-n} d u=d v$

$\Rightarrow u^{-n} d u=\frac{d v}{n-1}$

$\therefore \quad I=\int v^{\frac{1}{n}} \cdot \frac{d v}{n-1}=\frac{1}{n-1} \cdot \frac{v^{\frac{1}{n}+1}}{\frac{1}{n}+1}$

$=\frac{n}{n^{2}-1} v^{\frac{n+1}{n}}+C=\frac{n}{n^{2}-1}\left(1-\frac{1}{u^{n-1}}\right)^{\frac{n+1}{n}}+C$

$=\frac{n}{n^{2}-1}\left(1-\frac{1}{\sin ^{n-1} \theta}\right)^{\frac{n+1}{n}}+C$