Q. If $I_{n} = \int_{}^{}{\sin^{n}x\ dx}$,then $nI_{n} - (n - 1)I_{n - 2}$ equals
A) $\sin^{n - 1}{x\cos x}$
B) $\cos^{n - 1}{x\sin x}$
C) $- \sin^{n - 1}{x\cos x}$
D) $- \cos^{n - 1}{x\sin x}$
Explanation
(c)
We know that, if
$I_{n} = \int_{}^{}{\sin^{n}x\ dx}$, then
$I_{n} = - \frac{\sin^{n - 1}{x\cos x}}{n} + \frac{n + 1}{n}\ I_{n - 2}$
where $n$ is a positive integer
$\Rightarrow \ nI_{n} - (n - 1)I_{n + 2} = - \sin^{n - 1}{x\cos x}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.