Question

$\int_{0}^{2\pi}\theta sin^6\theta cos\theta\,d\theta$ is equal to

• A. $\frac{\pi}{6}$
• B. $\frac{3\pi}{16}$
• C. $\frac{16\pi}{3}$
• D. 0

Answer 1

Answer: (D)

0

Answer 2

Integral Equality : The equality 0 ∫ 2a f(x)dx = 2 0 ∫ a f(x)dx holds when the function f(x) satisfies the condition f(2a - x) = f(x).

Zero Integral Equivalence : The equation 0 ∫ 2a f(x)dx = 0 is valid if the function f(x) follows the pattern f(2a - x) = -f(x).

Given the integral I = 0 ∫ 2π a - bcosθ sin^2θ dθ, we observe that f(2π - θ) = a - bcos(2π - θ) / (2sin(2π - θ)cos(2π - θ)) simplifies to a - bcosθ / (-2sinθcosθ) = -f(θ).

Hence, it's established that I = 0 ∫ 2π a - bcosθ sin^2θ dθ = 0.

The correct answer is option (D): 0