Question

If \(I_n=\int \sin^n x\,dx\), then \(I_n-nI_{n-2}\) 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:

Use the standard reduction formula for \(\int \sin^n x\,dx\). Rearranging gives the required identity instantly.

Answer 1

The Correct Option is C

Step 1: Use reduction formula.
For \(\int \sin^n x\,dx\):
\[ I_n = -\frac{\sin^{n-1}x\cos x}{n} + \frac{n-1}{n}I_{n-2} \]
Step 2: Multiply by \(n\).
\[ nI_n = -\sin^{n-1}x\cos x + (n-1)I_{n-2} \]
Step 3: Rearrange to match required form.
\[ nI_n-(n-1)I_{n-2}=-\sin^{n-1}x\cos x \]
Thus:
\[ nI_n-(n-1)I_{n-2}=-\sin^{n-1}x\cos x \]
Final Answer:
\[ \boxed{-\sin^{n-1}x\cos x} \]