How do I integrate trigonometric functions raised to even powers?

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SUMMARY

This discussion focuses on integrating trigonometric functions raised to even powers, specifically the integrals of the form \int_{-\pi}^{\pi}\sin^{a}\theta\cos^b (2\theta) \ d\theta. The user seeks assistance with two specific integrals: \int_{-\pi}^{\pi}\sin^4\theta \ d\theta and \int_{-\pi}^{\pi}\sin^4\theta\cos(2\theta) \ d\theta. The user provides a breakdown of their attempts, including the use of the inner product and basis functions, but expresses confusion regarding the correct evaluation of these integrals. The discussion highlights the need for clarity in applying trigonometric identities and integration techniques.

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  • Understanding of trigonometric identities and functions
  • Familiarity with definite integrals and integration techniques
  • Knowledge of inner products in function spaces
  • Experience with manipulating and simplifying trigonometric expressions
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  • Study the evaluation of \int_{-\pi}^{\pi}\sin^4\theta \ d\theta using trigonometric identities
  • Learn techniques for integrating products of trigonometric functions, specifically \int_{-\pi}^{\pi}\sin^4\theta\cos(2\theta) \ d\theta
  • Explore the application of Fourier series in evaluating integrals of trigonometric functions
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Mathematicians, physics students, and anyone involved in advanced calculus or mathematical analysis, particularly those working with trigonometric integrals and Fourier analysis.

Dustinsfl
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Inner product:

\displaystyle <f,g>=\frac{1}{\pi}\int_{-\pi}^{\pi}fg \ dx=\begin{cases}0 & \ \text{if} \ f=g\\1 & \ \text{if} \ f\neq g\end{cases}

Basis:
\displaystyle\left\{\frac{1}{\sqrt{2}},\cos\theta, \sin\theta,\cdots\right\}

I am trying to remember how to integrals of the form:

\displaystyle \int_{-\pi}^{\pi}\sin^{a}\theta\cos^b (2\theta) \ d\theta

However, I getting no where.

I left some guidance with these two integrals and I should be good to go then.

\displaystyle\int_{-\pi}^{\pi}\sin^4\theta \ d\theta

\Rightarrow\int_{-\pi}^{\pi}\left(\frac{1}{2}-\frac{\cos(2\theta)}{2}\right)^2 \ d\theta

\Rightarrow \int_{-\pi}^{\pi}\left(\left(\frac{1}{\sqrt{2}}\right)^2-\frac{\cos(2\theta)}{2}\right)^2 \ d\theta

Now, I am drawing a blank.

The other one I need guidance on is:

\displaystyle\int_{-\pi}^{\pi}\sin^4\theta\cos(2\theta) \ d\theta

\Rightarrow\int_{-\pi}^{\pi}\left(\left(\frac{1}{\sqrt{2}}\right)^2-\frac{\cos(2\theta)}{2}\right)^2\cos(2\theta) \ d\theta

\Rightarrow\int_{-\pi}^{\pi}\frac{\cos(4\theta)\cos(2\theta)}{4} \ d\theta

Now I am stuck again.
 
Last edited:
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Dustinsfl said:
The other one I need guidance on is:

\displaystyle\int_{-\pi}^{\pi}\sin^4\theta\cos(2\theta) \ d\theta

\Rightarrow\int_{-\pi}^{\pi}\left(\left(\frac{1}{\sqrt{2}}\right)^2-\frac{\cos(2\theta)}{2}\right)^2\cos(2\theta) \ d\theta

\Rightarrow\int_{-\pi}^{\pi}\frac{\cos(4\theta)\cos(2\theta)}{4} \ d\theta

Now I am stuck again.

Bad math:

\int_{-\pi}^{\pi}\sin^4\theta\cos(2\theta) \ d\theta=
\int_{-\pi}^{\pi}\left[\frac{\cos(2\theta)}{4}-\frac{\cos^2(2\theta)}{2}+\frac{\cos(2\theta)* \cos^2(2\theta)}{4}\right] \ d\theta=
\int_{-\pi}^{\pi}\left[\frac{\cos(2\theta)}{4}-\left(\frac{1}{4}+\frac{\cos(4\theta)}{4}\right)+\frac{\cos(2\theta)}{8}+\frac{\cos(2\theta)*\cos(4\theta)}{8}\right] \ d\theta
(the Latex is correct so I don't know why it is all jacked up)

\int_{-\pi}^{\pi}\left[\frac{3\cos(2\theta)}{8}-\frac{1}{4}-\frac{\cos(4\theta)}{4}+\frac{\cos(2\theta)*\cos(4\theta)}{8}\right] \ d\theta
 
Can anyone provide any guidance?
 

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