Let $n\ge 2$ be a natural number and $0<\theta <\frac{\pi }{2}.$ Then, ${\displaystyle \underset{}{\overset{}{\int }}\frac{{\left({\sin }^n\theta -\sin \theta \right)}^{\frac{1}{n}}\cos \theta }{{\sin }^{n+1}\theta }d\theta }$ is equal to (where $C$ is a constant of integration)

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

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

$\therefore I={\displaystyle \underset{}{\overset{}{\int }}\frac{{\left(t^n=t\right)}^{1/n}}{t^{n+1}}dt}$

$={\displaystyle \underset{}{\overset{}{\int }}\frac{{\left[t^n\left(1-\frac{t}{t^n}\right)\right]}^{1/n}}{t^{n+1}}}$

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

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

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

$\Rightarrow \frac{dt}{t^n}=\frac{du}{n-1}$

$\Rightarrow t={\displaystyle \underset{}{\overset{}{\int }}\frac{u^{1/n}du}{n-1}}=\frac{u^{\frac{1}{n}+1}}{\left(n-1\right)\left(\frac{1}{n}+1\right)}+C$

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

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

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

$\left[\because u=1-\frac{1}{t^{n-1}}\text{and}k=\sin \theta \right]$