Question

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. \(-\cos^{n-1} x \sin x\)

Hint:

Reduction formula: \(\int \sin^n x dx = -\frac{\sin^{n-1} x \cos x}{n} + \frac{n-1}{n} \int \sin^{n-2} x dx\).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Use reduction formula for \(\int \sin^n x dx\).
Step 2: Detailed Explanation:
\(I_n = \int \sin^n x dx = -\frac{\sin^{n-1} x \cos x}{n} + \frac{n-1}{n} I_{n-2}\)
Multiply by \(n\): \(nI_n = -\sin^{n-1} x \cos x + (n-1)I_{n-2}\)
Thus \(nI_n - (n-1)I_{n-2} = -\sin^{n-1} x \cos x\)
Step 3: Final Answer:
\(-\sin^{n-1} x \cos x\).