Question

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

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

Answer 1

Answer: (C)

$\frac{1}{n-1}{(1-\frac{1}{{\sin }^{\pi -1}\theta })}^{\frac{\pi +1}{\pi }}+C$

$\int \frac{({\sin }^{\pi }\theta -\sin \theta {)}^{1/\pi }\cos \theta }{{\sin }^{\pi +1}\theta }d\theta$

$=\int \frac{\sin \theta {(1-\frac{1}{{\sin }^{\pi -1}\theta })}^{1/\pi }}{{\sin }^{\pi +1}\theta }d\theta$

Put $1-\frac{1}{{\sin }^{\pi -1}\theta }=t$

So $\frac{(n-1)}{{\sin }^{\pi }\theta }\cos \theta d\theta =dt$

Now $\frac{1}{n-1}\int (t{)}^{1/\pi }dt$

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

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