# Double integral

### Double integral

How to evaluate this double integral:

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

where $$A,B,C$$ are constants.
Guest

### Re: Double integral

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

Now change each $$sin^2(x)$$ to $$1- cos^2(x)$$:

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

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

See if you can do anything with that.

HallsofIvy

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