- #1

- 18

- 0

I'm trying to evaluate this integral over a hemisphere:

[tex]\int cos(\theta)^{a*cos^2(\phi) + b*sin^2(\phi)} dw[/tex]

where dw - solid angle measure,[tex]\phi[/tex] is azimuthal angle and [tex]\theta[/tex]

is polar angle.

Thus we have:

[tex]\int cos(\theta)^{a*cos^2(\phi) + b*sin^2(\phi)} dw[/tex] = [tex]\int \int cos(\theta)^{a*cos^2(\phi) + b*sin^2(\phi)} * sin(\theta) d\theta d\phi[/tex].

over hemishere.

Integral = [tex]\int^{2 * pi}_{0} \int^{pi/2}_{0}cos(\theta)^{a*cos^2(\phi) + b*sin^2(\phi)} * sin(\theta) d\theta d\phi[/tex]

evaluate inner integral:

making substitution u = cos([tex]\theta[/tex]),

[tex]\int^{pi/2}_{0}cos(\theta)^{a*cos^2(\phi) + b*sin^2(\phi)} * sin(\theta) d\theta =

- \int^{pi/2}_{0}u^{a*cos^2(\phi) + b*sin^2(\phi)} du = \frac{1}{a*cos^2(\phi) + b*sin^2(\phi) + 1}[/tex]

That's where I get stuck with integration wrt [tex]\phi[/tex]

Apparently this integral has to evaluate to [tex]\frac{1}{\sqrt{(a+1)(b+1)}}[/tex]