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

The Correct Option is D

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

Answer 2

To evaluate the integral \(\int_{0}^{2\pi}\theta \sin^6\theta \cos\theta\,d\theta\), we can use integration by parts.
Solution:
Let's denote \(u = \theta\) and \(dv = \sin^6\theta \cos\theta\,d\theta\). Then, we have:
\[ du = d\theta \]
\[ v = \int \sin^6\theta \cos\theta\,d\theta \]
Now, let's focus on finding \(v\). We can use a substitution to simplify the integral for \(v\).
Integration of \(v\):
Let \(t = \sin\theta\), then \(dt = \cos\theta\,d\theta\).
The integral becomes:
\[ v = \int t^6\,dt \]
\[ v = \frac{t^7}{7} + C \]
\[ v = \frac{\sin^7\theta}{7} + C \]
Applying Integration by Parts:
Now, we use the formula for integration by parts:
\[ \int u\,dv = uv - \int v\,du \]
Substitute \(u\), \(dv\), \(v\), and \(du\) into the formula:
\[ \int_{0}^{2\pi}\theta \sin^6\theta \cos\theta\,d\theta = \left[\theta \left(\frac{\sin^7\theta}{7}\right)\right]_{0}^{2\pi} - \int_{0}^{2\pi} \frac{\sin^7\theta}{7}\,d\theta \]
The first term evaluates to 0 because \(\sin(0) = \sin(2\pi) = 0\). We are left with:
\[ -\frac{1}{7} \int_{0}^{2\pi} \sin^7\theta\,d\theta \]
This integral can be simplified by using the identity \(\sin^2\theta = 1 - \cos^2\theta\), which can be iteratively applied to reduce the power of \(\sin\theta\) to an even number. However, the integral of \(\sin^{\text{odd}}\theta\) from 0 to \(2\pi\) is 0. Thus, the result of the integral is 0.
Final Answer:
\[ \int_{0}^{2\pi}\theta \sin^6\theta \cos\theta\,d\theta = 0 \]
So correct answer is option (D)