Let, n ≥ 2 be a natural number and 0 < θ < π 2 . Then ∫ s i n n θ - s i n…

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

Put, $\sin \theta =t\Rightarrow \cos \theta d\theta =dt$

$=\begin{array}{cc}\int \frac{{\left(t^n-t\right)}^{\frac{1}{n}}dt}{t^{n+1}}& \end{array}$

$=\int \frac{t{\left(1-\frac{1}{t^{n-1}}\right)}^{\frac{1}{n}}}{t^{n+1}}dt$

$= \int \frac{{\left(1-\frac{1}{t^{n-1}}\right)}^{\frac{1}{n}}}{t^n}dt$

Put $1-\frac{1}{t^{n-1}}=z$

$\Rightarrow \frac{\left(n-1\right)}{t^n}dt=dz$

$\Rightarrow I=\frac{1}{n-1}\int z^{\frac{1}{n}}dz$

Using $\int x^{n}dx=\frac{x^{n+1}}{n+1}+c$

$\Rightarrow I=\frac{z^{\frac{1}{n}+1}}{\left(\frac{1}{n}+1\right)\left(n-1\right)}+c$

$\Rightarrow I=\frac{n{\left(1-t^{1-n}\right)}^{\frac{1}{n}+1}}{n^2-1}+c,$ where $c$ is the constant of integration.
$\Rightarrow I=\frac{n}{n^2-1}{\left(1-\frac{1}{{\sin }^{n-1}\theta }\right)}^{\frac{n+1}{n}}+c.$