Q118 / 170 · Manipal (MET) 2019 · Mathematics
If $I_n=\int \sin ^n x d x$, then $n I_n-(n-1) I_{n-2}$ equals
- A$\sin ^{n-1} x \cos x$
- B$\cos ^{n-1} x \sin x$
- Correct answerC$-\sin ^{n-1} x \cos x$
- D$-\cos ^{n-1} x \sin x$
Solution
We know that, if $I_n=\int \sin ^n x d x$, then
$I_n=\frac{\sin ^{n-1} x \cos x}{n}+\frac{n-1}{n} I_{n-2}$
where $n$ is a positive integer.
$\Rightarrow n I_n-(n-1) I_{n-2}=-\sin ^{n-1} x \cos x$
Hence, the correct answer is option: 'C'.