Thompson scattering, simple integral?

[Kawada]

In theory this is a simpel integral problem which i can't solve

So I'm doing some plasma physics, and it comes with the derivation of the Thompson scattering (please bear with the first time I've tried using latex, I am sorry some of the greek lowercase laters look like superscripts, there not supposed to)
So I'm at the point where i have the total cross-section $$\sigma$$_T = the integral of (d$$\sigma$$/d$$\Omega$$) d$$\Omega$$.
ok so I've been given (d$$\sigma$$/d$$\Omega$$) = r_e²sin²$$\theta$$so the integral d$$\Omega$$ = d$$\theta$$d$$\phi$$, with $$\theta$$ between 0 and pi, and $$\phi$$ between 0 and 2pi.

but so the $$\phi$$ integral just gives 2pi, and in my attempt the integral of sin²$$\theta$$ with respect to $$\theta$$ over 0 and pi, is just pi/2?

yet looking at my notes, and the actual thompson scattering they have $$\sigma$$_T = 8pi*r_e²/3

now in my notes it says that the ingtegral over sin²$$\theta$$ d$$\theta$$ can become the integral over -sin²$$\theta$$ dcos$$\theta$$ still with $$\theta$$ between 0 and pi, and this yes, gives the required answer of 8pir_e²/3.

but hows do they change the intergral to that?

that is my only qualm, how they change the integral, and how my method of just ingetraing sin²$$\theta$$ d$$\theta$$ is not just pi/2?

i appreciate any help! :D (sorry for the long explanation)

 

[Kawada]

fail!

i've just worked it out! sorry, should i delete my original post? I am new to that is that easy to do? can an administrator easily do it?

the original integral shouldn't of been just d(theta)d(phi), but sin(theta)d(theta)d(phi), this makes the integral sin³(theta), and then when you change it to dcos(theta), you divide by -sin(theta)!

sorry about this! (also i really can't work this latex :S)