Question

If \( m \in \mathbb{Z}^+, n = 2m \) and \[ \int_0^{\frac{\pi}{2}} \sin^n x \cos^m x \, dx = K(m) \int_0^{\frac{\pi}{2}} \sin^m x \, dx, \text{ then } \frac{2^{n-1}(m - 1)!}{(2m - 1)!} \cdot K(m) = \]

• A. \( \frac{1}{m + 2} \cdot \frac{1}{m + 4} \cdot \ldots \cdot \frac{1}{m + r} \cdot \frac{1}{3m} \)
• B. \( \frac{1}{2m + 2} \cdot \frac{1}{2m + 4} \cdot \ldots \cdot \frac{1}{3m} \)
• C. \( \frac{\pi}{2} \cdot \frac{1}{m + 2} \cdot \frac{1}{m + 4} \cdot \ldots \cdot \frac{1}{m + r} \cdot \frac{1}{3m} \)
• D. \( \frac{\pi}{2} \cdot \frac{1}{2m + 2} \cdot \frac{1}{2m + 4} \cdot \ldots \cdot \frac{1}{3m} \)

Hint:

In problems involving integrals of the form \( \int \sin^n x \cos^m x \, dx \), use reduction formulas or symmetry when possible.

Answer 1

The Correct Option is A

This problem is based on properties of definite integrals involving powers of sine and cosine. The structure suggests a recursive or product form involving odd multiples and factorial simplifications. Recognizing the form of the reduction and substitution based on the given function and limits leads to a product expression for \( K(m) \) of the desired form.