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

To solve the problem involving the integral \(I_n = \int \sin^n x \, dx\) and find the expression for \(nI_n - (n - 1)I_{n-2}\), we will first perform integration by parts.

1. We start by considering the integration by parts formula: \(u \, dv = uv - \int v \, du\).
2. Let \(u = \sin^n x\) and \(dv = dx\). Therefore, \(du = n \sin^{n-1} x \cos x \, dx\) and \(v = x\).
3. Applying integration by parts: \[ I_n = \int \sin^n x \, dx = x \sin^n x - \int x n \sin^{n-1} x \cos x \, dx \] However, this leads to a more complex integral. Instead, we use a recursive method:
4. Use the reduction formula for powers of sine: \[ I_n = -\frac{1}{n} \sin^{n-1} x \cos x + \frac{n-1}{n} I_{n-2} \]
5. Substituting this into \(nI_n - (n - 1)I_{n-2}\), we have: \[ n \left(-\frac{1}{n} \sin^{n-1} x \cos x + \frac{n-1}{n} I_{n-2}\right) - (n-1)I_{n-2} \]
6. Simplifying the terms: \[ -\sin^{n-1} x \cos x + (n-1)I_{n-2} - (n-1)I_{n-2} = -\sin^{n-1} x \cos x \]
7. Hence, the simplified expression for \(nI_n - (n - 1)I_{n-2}\) is \(-\sin^{n-1} x \cos x\).

Therefore, the correct answer is \(-\sin^{n-1} x \cos x\).