Question

If \( I_{n} = \int \sin^{n} x \, dx \), then \( n I_{n} - (n-1) I_{n-2} \) equals

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

Hint:

This is the standard form of the reduction formula for $\sin^n x$.

Answer 1

The Correct Option is C

Step 1: Reduction Formula
The standard reduction formula for sine is $I_{n} = -\frac{\sin^{n-1}x \cos x}{n} + \frac{n-1}{n}I_{n-2}$.
Step 2: Rearrangement
Multiply by $n$: $nI_{n} = -\sin^{n-1}x \cos x + (n-1)I_{n-2}$. $\implies nI_{n} - (n-1)I_{n-2} = -\sin^{n-1}x \cos x$.
Final Answer: (c)