Double integral

[Guest]

How to evaluate this double integral:

[tex]\int_{x=0}^{x=\pi}\int_{y=0}^{y=2\pi}\frac{A^{2}B^{2}C^{2}\sin x}{\left(A^{2}B^{2}\cos^{2}x+A^{2}C^{2}\sin^{2}x\sin^{2}y+B^{2}C^{2}\sin^{2}x\cos^{2}y\right)^{3/2}}dydx[/tex]

where [tex]A,B,C[/tex] are constants.

[HallsofIvy]

First, change the order of integration:
[tex]A^2B^2C^2\int_{y= 0}^{2\pi}\int_{x= 0}^\pi \frac{sin(x)}{A^2B^2 cos^2(x)+ A^2C^2 sin^2(x)sin^2(y)+ B^2C^2 sin^2(x)cos^2(y)} dxdy[/tex].

Now change each [tex]sin^2(x)[/tex] to [tex]1- cos^2(x)[/tex]:

[tex]A^2B^2C^2\int_{y= 0}^{2\pi}\int_{x= 0}^\pi \frac{sin(x)}{A^2B^2 cos^2(x)+ A^2C^2 (1- cos^2(x))sin^2(y)+ B^2C^2 (1- cos^2(x))cos^2(y)} dxdy[/tex].

- Make the substitution u= cos(x) so that [tex]du= - sin(x)dx[/tex] and the integral becomes
[tex]A^2B^2C^2\int_{y= 0}^{2\pi}\int_{x= 0}^\pi \frac{-1}{A^2B^2 u^2+ A^2C^2 (1- u^2)sin^2(y)+ B^2C^2 (1- u^2)cos^2(y)} dudy[/tex]

See if you can do anything with that.