can someone solve this double integral step-by-step please???!!! It is from a differential solid angle problem: $$\int_{z=c}^{a+b}\int_{y=-b}^{b}\frac{x(dy)(dz)}{(x^2 + (2-y)^2 + (3.45-z)^2)^(3/2)] - [3*(x^2 + (2-y)^2 + (3.45-z)^2]}$$ (it is 3/2 squared!)
Thanks!!!
Observe that $$ \int \frac{1}{(t^2 + b)^{3/2}} \,\mathrm dt = \frac{t}{b\sqrt{b+t^2}} + \mathrm C $$ using the substitution $t = \sqrt{b} \tan s$ and $\mathrm dt = \sqrt{b} (\sec s)^2\,\mathrm ds$.
Repeative application of the above identity to the integrand gives you the result.