Question

\( \int_{-2 \pi}^{2 \pi} \sin ^{4} x \cos ^{6} x d x= \)

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

Hint:

For even powers of sine and cosine, the integral over a full period \( 0 \) to \( 2\pi \) is 4 times the integral over \( 0 \) to \( \pi/2 \).

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:

The integrand \( f(x) = \sin^4 x \cos^6 x \) is an even function and periodic with period \( \pi \). We can simplify the interval using symmetry properties and then use Wallis' Formula.
Step 2: Key Formula or Approach:

1. Periodicity: \( \int_{-a}^{a} f(x) dx = 2 \int_0^a f(x) dx \) (if even). 2. \( \int_0^{n\pi} f(x) dx = n \int_0^\pi f(x) dx \) (if period \( \pi \)). 3. Wallis Formula: \( \int_0^{\pi/2} \sin^m x \cos^n x dx = \frac{(m-1)!!(n-1)!!}{(m+n)!!} \cdot \frac{\pi}{2} \) (if m, n even).
Step 3: Detailed Explanation:

Limit is \( -2\pi \) to \( 2\pi \). Length \( 4\pi \). Since \( |\sin x| \) and \( |\cos x| \) repeat patterns every \( \pi/2 \) (powers are even), \( I = 2 \int_0^{2\pi} \sin^4 x \cos^6 x dx \) (Even func) \( I = 2 \times 4 \int_0^{\pi/2} \sin^4 x \cos^6 x dx = 8 I_W \). Using Wallis Formula for \( I_W \): \( I_W = \frac{(4-1)(2-1) \cdot (6-1)(4-1)(2-1)}{(10)(8)(6)(4)(2)} \cdot \frac{\pi}{2} \) \( I_W = \frac{3 \cdot 1 \cdot 5 \cdot 3 \cdot 1}{3840} \cdot \frac{\pi}{2} \) \( I_W = \frac{45}{3840} \frac{\pi}{2} = \frac{3}{256} \frac{\pi}{2} = \frac{3\pi}{512} \). Total Integral \( I = 8 \times \frac{3\pi}{512} = \frac{3\pi}{64} \).
Step 4: Final Answer:

Matches Option (D).