Question

If \( \int_{0}^{\pi/2} \sin^m x \cos^4 x \, dx = \frac{7\pi}{2048} \), then \( m = \)

• A. \( 8 \)
• B. \( 6 \)
• C. \( 10 \)
• D. \( 12 \)

Hint:

Integration Using Beta Functions}
Use Beta function for definite integrals of sine and cosine powers over \( [0, \pi/2] \).
Apply the identity \( B(p,q) = \frac{\Gamma(p)\Gamma(q)}{\Gamma(p+q)} \).
Remember that \( \Gamma(n+1) = n! \) and \( \Gamma(n + 1/2) \) involves square roots of \(\pi\).

Answer 1

The Correct Option is A

We are given the integral: \[ \int_0^{\pi/2} \sin^m x \cos^4 x \, dx = \frac{7\pi}{2048} \] Use the standard identity: \[ \int_0^{\pi/2} \sin^m x \cos^n x \, dx = \frac{1}{2} B\left(\frac{m+1}{2}, \frac{n+1}{2}\right) \] Here, \( n = 4 \Rightarrow \frac{n+1}{2} = \frac{5}{2} \) So, \[ \frac{1}{2} B\left(\frac{m+1}{2}, \frac{5}{2}\right) = \frac{7\pi}{2048} \] Using properties of Beta and Gamma functions: \[ B(p,q) = \frac{\Gamma(p)\Gamma(q)}{\Gamma(p+q)} \] Plug values and solve trial-wise for \( m = 8 \), we get: \[ B\left(\frac{9}{2}, \frac{5}{2}\right) = \frac{\Gamma(\frac{9}{2}) \Gamma(\frac{5}{2})}{\Gamma(7)} = \frac{105 \cdot \frac{3\sqrt{\pi}}{4}}{720} = \frac{7\pi}{1024} \Rightarrow \frac{1}{2} \cdot \frac{7\pi}{1024} = \frac{7\pi}{2048} \] Hence, \( \boxed{m = 8} \)