Definitite integrals involving sin and cos.

  • Thread starter tuite
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  • #1
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Hi!
I have learned how to solve equations like:
[tex]\int^{2\pi}_{0}\frac{1}{1+sin\theta}d\theta[/tex]

using

[tex]\int^{2\pi}_{0}F(sin\theta,cos\theta) d\theta[/tex]

[tex]\int_{C}F(\frac{z-z^{-1}}{2i},\frac{z+z^{-1}}{2}) \frac{dz}{iz}[/tex]

How do I solve equations of the type:
[tex]\int^{\pi}_{0}\frac{1}{1+sin\theta}d\theta[/tex]
and
[tex]\int^{\pi}_{-\pi}\frac{1}{1+sin\theta}d\theta[/tex]

Please help :-)
 

Answers and Replies

  • #2
gabbagabbahey
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How do I solve equations of the type:
[tex]\int^{\pi}_{0}\frac{1}{1+sin\theta}d\theta[/tex]
and
[tex]\int^{\pi}_{-\pi}\frac{1}{1+sin\theta}d\theta[/tex]

Please help :-)

The second one is reallly the exact same type as the one you know how to solve. To see this, just make the substitution [tex]\overbar{\theta}=\theta-\pi[/itex]

As for the first one, draw a picture....Using [itex]z=e^{i\theta}[/itex], you should see that when you integrate from [itex]0[/itex] to [itex]\pi[/itex] you are integrating over the semi-circular arc in the upper half plane instead of integrating over the entire unit circle. So, the result is that only residues of the poles that lie inside the unit circle and are in the upper half-plane will contribute to the result.
 
  • #3
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0
The second one is reallly the exact same type as the one you know how to solve. To see this, just make the substitution [tex]\overbar{\theta}=\theta-\pi[/itex]

As for the first one, draw a picture....Using [itex]z=e^{i\theta}[/itex], you should see that when you integrate from [itex]0[/itex] to [itex]\pi[/itex] you are integrating over the semi-circular arc in the upper half plane instead of integrating over the entire unit circle. So, the result is that only residues of the poles that lie inside the unit circle and are in the upper half-plane will contribute to the result.

Ah! Great! The substitution explained for me! Thanks!

Right now I'm struggling with:
[tex]\int^{\pi}_{-\pi}\frac{1}{1+sin^{2}{\theta}}d\theta[/tex]

when rewritten:

[tex]\int_{C}\frac{1}{1+(\frac{z-z^{-1}}{2i})^2}\frac{dz}{iz} [/tex]
[tex]\int_{C}\frac{4}{4-(z^2-2+\frac{1}{z^2})}\frac{dz}{iz} [/tex]

which I get to

[tex]\int_{C}\frac{4}{(6z-z^3-\frac{1}{z})i}dz[/tex]

and then I'm stuck.
Any hints?
 
  • #4
gabbagabbahey
Homework Helper
Gold Member
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Multiply both the numerator and denominator by [itex]iz[/itex] and then factor the resulting denominator (use the quadratic equation)
 
  • #5
1,838
7
The semi-contour is not a closed contour. That has to be fixed first before you can aply the residue theorem.
 
  • #6
34,875
6,613
Hi!
I have learned how to solve equations like:
[tex]\int^{2\pi}_{0}\frac{1}{1+sin\theta}d\theta[/tex]

using

[tex]\int^{2\pi}_{0}F(sin\theta,cos\theta) d\theta[/tex]
A small point: You aren't "solving equations." You are evaluating definite integrals. An equation has an = in it.
 

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