Question

Let $n\ge 2$ be a natural number and $0<\theta <p/2$.
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 $C$ is a constant of integration)

• A. $\frac{n}{n^2-1}{\left(1-\frac{1}{{\sin }^{n+1}\theta }\right)}^{\frac{n+1}{n}}+C$
• B. $\frac{n}{n^2+1}{\left(1-\frac{1}{{\sin }^{n-1}\theta }\right)}^{\frac{n+1}{n}}+C$
• C. $\frac{n}{n^2-1}{\left(1-\frac{1}{{\sin }^{n-1}\theta }\right)}^{\frac{n+1}{n}}+C$
• D. $\frac{n}{n^2-1}{\left(1+\frac{1}{{\sin }^{n-1}\theta }\right)}^{\frac{n+1}{n}}+C$

Answer 1

Answer: (C)

$\int \frac{{\left({\sin }^n\theta -\sin \theta \right)}^{1/n}\cos \theta }{{\sin }^{n+1}\theta }d\theta$
$=\int \frac{\sin \theta {\left(1-\frac{1}{{\sin }^{n-1}\theta }\right)}^{1/n}}{{\sin }^{n+1}\theta }d\theta$
Put $1-\frac{1}{{\sin }^{n-1}\theta }=t$
So $\frac{(n-1)}{{\sin }^n\theta }\cos \theta d\theta =dt$
Now $\frac{1}{n-1}\int (t{)}^{1/n}dt$
$=\frac{1}{(n-1)}\frac{(t{)}^{\frac{1}{n}+1}}{\frac{1}{n}+1}+C$
$\frac{1}{(n-1)}{\left(1-\frac{1}{{\sin }^{n-1}\theta }\right)}^{\frac{1}{n}+1}+C$