Question

The value of the integral \( \int_{-2\pi}^{2\pi} \sin^4 x \cos^6 x \, dx \) is equal to:

• A. \( \frac{3\pi}{128} \)
• B. \( \frac{9\pi}{32} \)
• C. \( \frac{9\pi}{64} \)
• D. \( \frac{3\pi}{64} \)

Hint:

For \( \int_{0}^{n\pi} \sin^m x \cos^n x \, dx \), if \( m, n \) are even, the integral is \( 2n \times \int_{0}^{\pi/2} \). For \( \int_{-2\pi}^{2\pi} \), that is 4 full periods, or 8 half-periods.

Answer 1

The Correct Option is C

Concept: This problem combines symmetry properties of definite integrals with Wallis' Formula:
• Even Function Property: If \( f(-x) = f(x) \), then \( \int_{-a}^{a} f(x) \, dx = 2 \int_{0}^{a} f(x) \, dx \).
• Periodic Property: If \( f(x) \) is periodic with period \( T \), then \( \int_{0}^{nT} f(x) \, dx = n \int_{0}^{T} f(x) \, dx \).
• Wallis' Formula: Used for \( \int_{0}^{\pi/2} \sin^m x \cos^n x \, dx \).

Step 1: Using symmetry and periodicity to reduce limits.
The integrand \( f(x) = \sin^4 x \cos^6 x \) is an even function. Therefore: \[ I = 2 \int_{0}^{2\pi} \sin^4 x \cos^6 x \, dx \] The function \( \sin^4 x \cos^6 x \) is periodic with period \( \pi \). However, inside the squares/even powers, the behavior repeats every \( \pi/2 \) in terms of area under the curve. Specifically, \( \int_{0}^{2\pi} = 4 \int_{0}^{\pi/2} \): \[ I = 2 \times 4 \int_{0}^{\pi/2} \sin^4 x \cos^6 x \, dx = 8 \int_{0}^{\pi/2} \sin^4 x \cos^6 x \, dx \]

Step 2: Applying Wallis' Formula.
For \( m=4, n=6 \), both are even, so we multiply by \( \frac{\pi}{2} \): \[ \int_{0}^{\pi/2} \sin^4 x \cos^6 x \, dx = \frac{(3 \cdot 1) \cdot (5 \cdot 3 \cdot 1)}{(10 \cdot 8 \cdot 6 \cdot 4 \cdot 2)} \times \frac{\pi}{2} \] \[ = \frac{45}{3840} \times \frac{\pi}{2} = \frac{3}{256} \times \frac{\pi}{2} = \frac{3\pi}{512} \]

Step 3: Final Calculation.
Substitute the value back into the expression from
Step 1: \[ I = 8 \times \left( \frac{3\pi}{512} \right) = \frac{3\pi}{64} \]