Definitite integrals involving sin and cos.

[tuite]

Hi!
I have learned how to solve equations like:
\int^{2\pi}_{0}\frac{1}{1+sin\theta}d\theta

using

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

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

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

Please help :-)

 

[gabbagabbahey]

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

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 \overbar{\theta}=\theta-\pi[/itex]<br /> <br /> As for the first one, draw a picture...Using z=e^{i\theta}, you should see that when you integrate from 0 to \pi 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.

 

[tuite]

The second one is reallly the exact same type as the one you know how to solve. To see this, just make the substitution \overbar{\theta}=\theta-\pi[/itex]<br /> <br /> As for the first one, draw a picture...Using z=e^{i\theta}, you should see that when you integrate from 0 to \pi 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.

<br /> <br /> Ah! Great! The substitution explained for me! Thanks!<br /> <br /> Right now I&#039;m struggling with:<br /> \int^{\pi}_{-\pi}\frac{1}{1+sin^{2}{\theta}}d\theta<br /> <br /> when rewritten:<br /> <br /> \int_{C}\frac{1}{1+(\frac{z-z^{-1}}{2i})^2}\frac{dz}{iz}<br /> \int_{C}\frac{4}{4-(z^2-2+\frac{1}{z^2})}\frac{dz}{iz}<br /> <br /> which I get to<br /> <br /> \int_{C}\frac{4}{(6z-z^3-\frac{1}{z})i}dz<br /> <br /> and then I&#039;m stuck. <br /> Any hints?

 

[gabbagabbahey]

Multiply both the numerator and denominator by iz and then factor the resulting denominator (use the quadratic equation)

 

[Count Iblis]

The semi-contour is not a closed contour. That has to be fixed first before you can aply the residue theorem.

 

[Mark44]

Hi!
I have learned how to solve equations like:
\int^{2\pi}_{0}\frac{1}{1+sin\theta}d\theta

using

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

A small point: You aren't "solving equations." You are evaluating definite integrals. An equation has an = in it.