# An integral

1. Nov 21, 2006

### quasar987

Does anyone know how to do this integral? Mapple just thinks and think and never gives an answer. I couldn't find a primitive in a table either.

$$\int_{0}^{\pi}\int_{0}^{+\infty}\frac{r^4\sin^3\theta}{(c+br\cos\theta+(\sin^2\theta )r^2)^3}drd\theta$$

In one attempt, I used a table to reduce the r integral to

$$\int_{0}^{\pi}\int_0^{+\infty}\frac{\sin^3\theta}{(c+br\cos\theta+(\sin^2\theta) r^2)^3}dr$$

But what's the integral of that?

Last edited: Nov 21, 2006
2. Nov 21, 2006

### arildno

Complete the square within the denominator with rcos(theta) as one of your terms within that square.
Then, make a change of variables from theta to cos(theta) (i.e, integrate from 1 to -1); probably, you need to use partial fractions decomposition.

You'll end up (hopefully!) with a rational function in r, which you then are to integrate with respect to r.

Last edited: Nov 21, 2006