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:

For integrals of the form \(\int_{0}^{\pi/2}\sin^{2p}x\cos^{2q}x\,dx\), use the double factorial formula to quickly evaluate the result.

Answer 1

The Correct Option is A

Step 1: Check the value from the options.
We need \[ \int_{0}^{\pi/2}\sin^m x\cos^4 x\,dx=\frac{7\pi}{2048} \] The correct option is likely an even value of \(m\), because integrals of even powers of sine and cosine over \(\left[0,\frac{\pi}{2}\right]\) often contain \(\pi\).

Step 2: Test \(m=8\).
For \(m=8\), the integral becomes \[ \int_{0}^{\pi/2}\sin^8 x\cos^4 x\,dx \] Using the standard result, \[ \int_{0}^{\pi/2}\sin^{2p}x\cos^{2q}x\,dx = \frac{(2p-1)!!(2q-1)!!}{(2p+2q)!!}\cdot \frac{\pi}{2} \] Here, \[ 2p=8 \Rightarrow p=4 \] and \[ 2q=4 \Rightarrow q=2 \] So, \[ \int_{0}^{\pi/2}\sin^8 x\cos^4 x\,dx = \frac{7!!\cdot 3!!}{12!!}\cdot \frac{\pi}{2} \]

Step 3: Simplify the double factorials.
Now, \[ 7!!=7\cdot 5\cdot 3\cdot 1=105 \] and \[ 3!!=3\cdot 1=3 \] Also, \[ 12!!=12\cdot 10\cdot 8\cdot 6\cdot 4\cdot 2 \] \[ 12!!=46080 \] Therefore, \[ \int_{0}^{\pi/2}\sin^8 x\cos^4 x\,dx = \frac{105\cdot 3}{46080}\cdot \frac{\pi}{2} \] \[ = \frac{315}{46080}\cdot \frac{\pi}{2} \] \[ = \frac{315\pi}{92160} \] Simplifying, \[ \frac{315\pi}{92160}=\frac{7\pi}{2048} \]

Step 4: Match with the given value.
The given value is \[ \frac{7\pi}{2048} \] This matches exactly when \[ m=8 \]

Step 5: Final conclusion.
Therefore, \[ \boxed{8} \]