Integration homework

  • #1

Homework Statement



Find an expression for

[tex]\cos 2\theta + \cos 4\theta + \cos 6\theta + \dots + \cos (2n\theta) [/tex]

Hence prove that
[tex]\int_0^{\pi/2} \frac{\sin(2n+1)x}{\sin x} \ dx = \frac{\pi}{2} [/tex]


Homework Equations



[tex]\cos 2\theta + \cos 4\theta + \cos 6\theta + \dots + \cos (2n\theta) [/tex]
and
[tex]\int_0^{\pi/2} \frac{\sin(2n+1)x}{\sin x} \ dx = \frac{\pi}{2} [/tex]

The Attempt at a Solution



I found an expression for [tex]\cos 2\theta + \cos 4\theta + \cos 6\theta + \dots + \cos (2n\theta) [/tex] which was [tex]\sum_{k=1}^{n}1-2sin^{2}k\theta[/tex] but couldn't continue because it doesn't look like the appropriate expression.
 
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Answers and Replies

  • #2
SammyS
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Homework Statement



Find an expression for

[tex]\cos 2\theta + \cos 4\theta + \cos 6\theta + \dots + \cos (2n\theta) [/tex]

Hence prove that
[tex]\int_0^{\pi/2} \frac{\sin(2n+1)x}{\sin x} \ dx = \frac{\pi}{2} [/tex]


Homework Equations



[tex]\cos 2\theta + \cos 4\theta + \cos 6\theta + \dots + \cos (2n\theta) [/tex]
and
[tex]\int_0^{\pi/2} \frac{\sin(2n+1)x}{\sin x} \ dx = \frac{\pi}{2} [/tex]

The Attempt at a Solution



I found an expression for [tex]\cos 2\theta + \cos 4\theta + \cos 6\theta + \dots + \cos (2n\theta) [/tex] which was [tex]\sum_{k=1}^{n}1-sin^{2}k\theta[/tex] but couldn't continue because it doesn't look like the appropriate expression.
What is 1 - sin2(u) ?
 
  • #3
oh it should have been 1-2sin^2u
edited*
 
  • #4
SammyS
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  • #5
cos(2u) = ?

Yes, that is how I go 1-2sin^2(u) in the first place (because the integral had sin) but I don't see how it can help me evaluate the integral :(
 
  • #6
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Do you know the identity that [itex]cos θ= \frac{e^{iθ} + e^{-iθ}}{2}[/itex]?
 
  • #7
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It seems helpful to use this identity to reduce the question to geometric progression, then what it takes is some computation.
 
  • #8
vela
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Along the same lines as raopeng's hint, but a little bit less tedious:

##\cos 2x + \cos 4x + \cdots + \cos (2nx)## is equal to the real part of ##e^{2ix} + e^{4ix} + \cdots + e^{2nix}##
 
  • #10
vela
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If you calculate the sum correctly, it'll become clear.
 
  • #11
I have never learnt how to express cis in terms of e so can you please give me a hint on how to do it?
Thank you!
 
  • #12
vela
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Not really sure what kind of hint you're looking for.
 
  • #14
vela
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Use Euler's formula, keeping in mind that ##x## is real.
 
  • #15
I get
e^2ix=cos2x+ison2x
e^4ix=cos4x+ison4x
.
.
.
e^nix=cosnx+isonnx

I still don't see how this can help with the integral :(
 
  • #16
vela
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Hint: ##e^{2ix} + e^{4ix} + \cdots + e^{2nix} = (e^{2ix})^1 + (e^{2ix})^2 + \cdots + (e^{2ix})^n##. You have a geometric progression on your hands, as raopeng noted earlier.
 
  • #17
Ok, so I get [tex]\frac{e^{ix}+e^{-ix}}{2} =cosx [/tex]
[tex]\frac{e^{ix}-e^{-ix}}{2} =sinx[/tex]
 
  • #19
vela
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We're not here to spoon-feed you the solution. You need to show some initiative at solving the problem yourself.
 
  • #20
I know this is dumb but this is what I did :(
[tex]\int_{0}^{\frac{\pi }{2}}\frac{e^{i(2n+1)x}-e^{-i(2n+1)x}}{e^{ix}-e^{-ix}}dx[/tex]
 
  • #21
I know, you are not allowed to solve the question form me but if the student is stuck shouldn't you give some hints? :(
 
  • #22
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It is already very close! Remember we have the identity that [itex]sinθ=\frac{e^iθ-e^{-iθ}}{2i}[/itex], how does this relate to the integration you wrote?

BTW if you find such problems difficult, given that you haven't learned about Euler's Equation, it is perhaps more productive to have a better grasp of the material. It is indeed better than asking as it will improve your understanding on the topic a lot more. And all the joy of revelation after solving a problem will be the greatest when you figure it out completely on your own :)
 
  • #23
The denominator is equal to 2isinx and the numerator =2isin(2n+1)x?
How can I integrate the integral I have in post 20 though, it looks very hard :(
 
  • #24
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yes and it would equal integrating a sum of cos because you just prove that these two expressions are equivalent. And integrating separately the cos in the series is an easy job.
 
  • #25
But why is
[tex]\int_{0}^{\frac{\pi }{2}}\frac{e^{i(2n+1)x}-e^{-i(2n+1)x}}{e^{ix}-e^{-ix}}dx[/tex]
equal to the cos2x series? Isn't this the imaginary part of e^2ix+e^4ix+...+e^2nix and the cos2x series the real part?
 
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