What is the integral of sin²(kθ)cos²(kθ) over the interval from 0 to 2π?

Click For Summary

Homework Help Overview

The discussion revolves around evaluating the integral of sin²(kθ)cos²(kθ) over the interval from 0 to 2π, exploring trigonometric identities and integration techniques.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the use of trigonometric identities, particularly sin(2x), to simplify the integral. There is mention of substitution attempts that did not yield results. One participant expresses concern about the simplicity of the resulting answer and questions whether k should still be present in the final expression.

Discussion Status

The discussion is active, with participants providing insights into potential simplifications and questioning the implications of k being an integer or not. There is no explicit consensus on the final evaluation of the integral, and multiple interpretations are being explored.

Contextual Notes

Participants are considering the implications of k being an integer and how it affects the outcome of the integral, indicating a potential constraint in the problem setup.

bigplanet401
Messages
101
Reaction score
0

Homework Statement


What is

[tex] \int_0^{2 \pi} \; d\theta \sin^2 k\theta \cos^2 k\theta \; ?[/tex]

Homework Equations


Orthogonality of sines and cosines?

The Attempt at a Solution


I tried substitution and didn't get anywhere. Yeah, that's about it.
 
Physics news on Phys.org
Use the trig identity sin(2x)=2*sin(x)*cos(x) for a start.
 
Whoa, I completely missed that. Using that identity, the integral becomes

[tex] \begin{align*}<br /> & \int_0^{2\pi} d\theta \; \frac{1}{4} \sin^2 2k\theta\\<br /> &= \int_0^{2\pi} d\theta \; \frac{1}{8} (1 - \sin 4 k \theta )\\<br /> &= \frac{\pi}{4} \, ,<br /> \end{align*}[/tex]
right?

The answer just seems too simple--like there should be some k's around or something.
 
You mean 1-cos(4k*theta), I hope. If k is an integer then there are no k's left around. If it isn't there are.
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 17 ·
Replies
17
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
4K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K