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 Symmetry and Evaluation 

Integral Equality: The equality $ \int_0^{2a} f(x)\,dx = 2 \int_0^{a} f(x)\,dx $ holds when the function satisfies $ f(2a - x) = f(x) $.

Zero Integral Equivalence: The integral $ \int_0^{2a} f(x)\,dx = 0 $ if the function satisfies $ f(2a - x) = -f(x) $.

Given: $ I = \int_0^{2\pi} \frac{a - b\cos\theta}{2\sin\theta\cos\theta} \, d\theta $

To test symmetry, evaluate $ f(2\pi - \theta) $:

$ f(2\pi - \theta) = \frac{a - b\cos(2\pi - \theta)}{2\sin(2\pi - \theta)\cos(2\pi - \theta)} = \frac{a - b\cos\theta}{-2\sin\theta\cos\theta} = -f(\theta) $

Conclusion: Since $ f(2\pi - \theta) = -f(\theta) $, the function is odd about $ \pi $, and the integral over $ [0, 2\pi] $ is therefore:

$ I = 0 $

Final Answer: Option (D): 0

Answer 2

Evaluation of the Integral 

Given: $ \int_{0}^{2\pi} \theta \sin^6\theta \cos\theta \, d\theta $

Step 1: Integration by Parts

Let $ u = \theta $ and $ dv = \sin^6\theta \cos\theta\, d\theta $. Then:

$ du = d\theta $

We now compute $ v = \int \sin^6\theta \cos\theta \, d\theta $

Step 2: Substitution

Let $ t = \sin\theta $, then $ dt = \cos\theta\, d\theta $

So the integral becomes:

$ v = \int t^6\, dt = \frac{t^7}{7} + C = \frac{\sin^7\theta}{7} + C $

Step 3: Apply Integration by Parts Formula

$ \int u\, dv = uv - \int v\, du $

So,

$ \int_{0}^{2\pi} \theta \sin^6\theta \cos\theta \, d\theta = \left[\theta \cdot \frac{\sin^7\theta}{7} \right]_{0}^{2\pi} - \int_{0}^{2\pi} \frac{\sin^7\theta}{7} \, d\theta $

Step 4: Evaluate the Boundary Terms

$ \sin(0) = \sin(2\pi) = 0 $, so the first term becomes 0.

Thus the integral becomes:

$ -\frac{1}{7} \int_{0}^{2\pi} \sin^7\theta \, d\theta $

Step 5: Analyze the Integral

Since $ \sin^7\theta $ is an odd power of the sine function, and sine is periodic with zero average over $ [0, 2\pi] $, we have:

$ \int_{0}^{2\pi} \sin^7\theta \, d\theta = 0 $

Final Answer:

$ \int_{0}^{2\pi} \theta \sin^6\theta \cos\theta \, d\theta = 0 $

Correct option: (D)