Let $\mathrm{n} \geq 2$ be a natural number and $0<\theta<\pi / 2$.
Then $\int \frac{\left(\sin ^{\mathrm{n}} \theta-\sin \theta\right)^{\frac{1}{n}} \cos \theta}{\sin ^{\mathrm{n}+1} \theta} \mathrm{d} \theta$ is equal to :
(Where $\mathrm{C}$ is a constant of integration)
Correct Option: , 3
$\int \frac{\left(\sin ^{n} \theta-\sin \theta\right)^{1 / n} \cos \theta}{\sin ^{n+1} \theta} d \theta$
$=\int \frac{\sin \theta\left(1-\frac{1}{\sin ^{\mathrm{n}-1} \theta}\right)^{1 / \mathrm{n}}}{\sin ^{\mathrm{n}+1} \theta} \mathrm{d} \theta$
Put $1-\frac{1}{\sin ^{n-1} \theta}=\mathrm{t}$
So $\frac{(n-1)}{\sin ^{n} \theta} \cos \theta d \theta=d t$
Now $\frac{1}{n-1} \int(t)^{1 / n} d t$
$=\frac{1}{(n-1)} \frac{(t)^{\frac{1}{n}+1}}{\frac{1}{n}+1}+C$
$=\frac{1}{(n-1)}\left(1-\frac{1}{\sin ^{n-1} \theta}\right)^{\frac{1}{n}+1}+C$