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

Tri Integration

  • Thread starter icystrike
  • Start date
  • #1
446
1

Homework Statement


Given [tex]\int_{0}^{k\pi} cos^{2m}(\theta) d\theta = A [/tex]
Express [tex]\int_{0}^{k\pi} cos^{2m}(\theta) cos(2\theta) [/tex] in terms of A.



Totally Stucked .. :X


I've substituted [tex]cos^{2}\theta = \frac{1+cos(2\theta)}{2}[/tex]

Then i get 1/2 A + another chunck of integral.

I've used integration by parts to tackle the chunck.

such that i integrate the [tex]cos(2\theta)[/tex].

I've gotten pretty good simplification but i think i've made some mistakes here a there.

Homework Equations





The Attempt at a Solution

 
Last edited:

Answers and Replies

  • #2
124
0
A method you could try is to remember that [tex]\cos (2 \theta) = 2 \cos ^2 (\theta) - 1[/tex]. The second integral thus becomes

[tex]\int_0^{k \pi}\cos ^{2m} (\theta)(2 \cos ^2 (\theta) - 1) d \theta = 2 \int_0^{k \pi}\cos ^{2m+2} (\theta) d \theta - A,[/tex]​

and you could try to use a http://en.wikipedia.org/wiki/Integration_by_reduction_formulae" [Broken] on this last integral to try to get it as a function of A.

Hope this helps. :)
 
Last edited by a moderator:
  • #3
446
1
But once agn we are stucked with the integral of [tex]cos^{2m+2}\theta[/tex]

:)
 
  • #4
124
0
...have you tried looking at the reduction formulas on that Wikipedia page I linked you to?
 
  • #5
446
1
...have you tried looking at the reduction formulas on that Wikipedia page I linked you to?
e91bac2435414ec2e44413cee3945b65.png


Are u referring to this? Seems to be great! Havent tried though...
 
  • #6
124
0
Precisely the one! Remember that for us, our [tex]n[/tex] is [tex]2m + 2[/tex] and that since we have a definite integral, it just so happens that [tex]\sin(x)[/tex] vanishes for 0 and [tex]k \pi[/tex], where [tex]k \in \mathbb{Z}[/tex] ... :)
 
  • #7
446
1
Precisely the one! Remember that for us, our [tex]n[/tex] is [tex]2m + 2[/tex] and that since we have a definite integral, it just so happens that [tex]\sin(x)[/tex] vanishes for 0 and [tex]k \pi[/tex], where [tex]k \in \mathbb{Z}[/tex] ... :)
Fit perfectly well! Thanks a million! :rofl:
 

Related Threads on Tri Integration

  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
3
Views
806
  • Last Post
Replies
6
Views
1K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
3
Views
961
  • Last Post
Replies
2
Views
786
Replies
2
Views
1K
Replies
6
Views
1K
Replies
13
Views
825
Replies
3
Views
1K
Top