Question

If $I_n=\int {\sin }^{n}xdx,n\in N,n\ge 2$ then $n{I}_n-(n-1)I_{n-2}=$
(where $C$ is integration constant)

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

Answer 1

The correct option is A $-\cos x\cdot {\sin }^{n-1}x+C$

We know that,
$I_n=\int {\sin }^{n}xdx=-\frac{\cos x{\sin }^{n-1}x}{n}+\frac{(n-1)}{n}{I}_{n-2}$
$\Rightarrow n{I}_n-(n-1)I_{n-2}=-\cos x\cdot {\sin }^{n-1}x+C$