• Support PF! Buy your school textbooks, materials and every day products Here!

Integration (inner product)

  • Thread starter Dustinsfl
  • Start date
  • #1
699
5
Inner product:

[itex]\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}[/itex]

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

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

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

However, I getting no where.

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

[itex]\displaystyle\int_{-\pi}^{\pi}\sin^4\theta \ d\theta[/itex]

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

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

Now, I am drawing a blank.

The other one I need guidance on is:

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

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

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

Now I am stuck again.
 
Last edited:

Answers and Replies

  • #2
699
5
The other one I need guidance on is:

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

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

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

Now I am stuck again.
Bad math:

[tex]\int_{-\pi}^{\pi}\sin^4\theta\cos(2\theta) \ d\theta=[/tex]
[tex]\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=[/tex]
[tex]\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[/tex]
(the Latex is correct so I don't know why it is all jacked up)

[tex]\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[/tex]
 
  • #3
699
5
Can anyone provide any guidance?
 

Related Threads on Integration (inner product)

  • Last Post
Replies
1
Views
2K
Replies
1
Views
5K
Replies
5
Views
3K
  • Last Post
Replies
1
Views
826
  • Last Post
Replies
6
Views
875
  • Last Post
Replies
3
Views
951
  • Last Post
Replies
10
Views
2K
  • Last Post
Replies
2
Views
864
  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
8
Views
1K
Top