Q77 / 120 · VITEEE 2009 · 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
$$
\begin{array}{l}
I_{n}=\int \sin ^{n} x d x, \text { then } \\
I_{n}=-\frac{\sin ^{n-1} x \cos x}{n}+\frac{n-1}{n} I_{n-2}
\end{array}
$$
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'.