Question

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

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

Answer 1

The correct option is A $\frac{n}{n^2-1}{(1-\frac{1}{{\sin }^{n-1}\theta })}^{\frac{n+1}{n}}+C$

$I=\int \frac{{({\sin }^n\theta -\sin \theta )}^{\frac{1}{n}}\cos \theta }{{\sin }^{n+1}\theta }d\theta$
$=\int \frac{\sin \theta {(1-{\sin }^{1-n}\theta )}^{\frac{1}{n}}\cos \theta }{{\sin }^{n+1}\theta }d\theta$
$=\int \frac{{(1-{\sin }^{1-n}\theta )}^{\frac{1}{n}}\cos \theta }{{\sin }^n\theta }d\theta$
Take $1-{\sin }^{1-n}\theta =t$
$-(1-n)\cdot {\sin }^{-n}\theta \cdot \cos \theta d\theta =dt$
$\therefore I=\frac{1}{n-1}\int t^{\frac{1}{n}}dt$
$=\frac{1}{n-1}\cdot \frac{t^{\frac{1}{n}+1}}{\frac{1}{n}+1}+C$
$=\frac{n}{n^2-1}{(1-{\sin }^{1-n}\theta )}^{\frac{n+1}{n}}+C$
$=\frac{n}{n^2-1}{(1-\frac{1}{{\sin }^{n-1}\theta })}^{\frac{n+1}{n}}+C$