Question

If $I_n=\int {\sin }^{n}xdx,$ then $n{I}_n-\left(n-1\right)I_{n-2}$ equals

• A. ${\sin }^{n-1}x\cos x$
• B. ${\cos }^{n-1}x\sin x$
• C. $-{\sin }^{n-1}x\cos x$
• D. $-{\cos }^{n-1}x\sin x$

Answer 1

Answer: (C)

We know that, if
$I_n=\int {\sin }^{n}xdx,$ then
$I_n=-\frac{{\sin }^{n-1}x\cos x}{n}+\frac{n-1}{n}{I}_{n-2}$
where $n$ is a positive integer.
$\Rightarrow n{I}_n-(n-1)I_{n-2}=-{\sin }^{n-1}x\cos x$