I was reading this article: https://webspace.science.uu.nl/~maas0131/files/MaasJCAM1994.pdf
I came to the equation (24):
The author says... "We thus find, by the same argument,":
$$\int_0^1 f(\xi)E(k\sqrt{1-\xi^2})d\xi=\int_0^1 \sqrt{1-k^2\xi^2}\int_0^{\frac{\pi}{2}}f(\sqrt{1-\xi^2}cos\phi)d\phi d\xi$$
I can not prove it. Can you help me?
I am also stack in a line of the same article. He wrote (below the equation (49)): $$\int_0^{\pi/2}\frac{1}{(1+\rho\cos^2\phi)^2}d\phi=\frac{1}{4}\pi\frac{2+\rho}{(1+\rho)^{3/2}}$$ How can I do that? Thanks.
Fundamentally, what we want to show is that $$ \int_0^1\!\!\!\int_0^{\pi/2}f(\xi)\sqrt{1-k^2(1-\xi^2)\sin^2\phi} d\phi d\xi = \int_0^1\!\!\!\int_0^{\pi/2}f\left(\sqrt{1-u^2}\cos\theta\right)\sqrt{1-k u^2}d\theta du, $$ where I have renamed the dummy variables in the second integral to reduce ambiguity. It turns out the direct change of variables $$ \xi \rightarrow \sqrt{1-u^2}\cos\theta\,\,\,,\,\,\,\sqrt{1-\xi^2}\sin\phi \rightarrow u. $$ has a Jacobian of $1$. To see this, first show that $\sqrt{1-\xi^2}\cos\phi d\phi d\xi = \sqrt{1-u^2}\sin\theta d \theta d u$ by differentiating both sides of the change of variable equations. Then we use $$ (1-\xi^2)\cos\phi^2 = (1-\xi^2) - (1-\xi^2)\sin^2\phi = 1-(1-u^2)\cos^2\theta - u^2 = (1-u^2)\sin^2\theta. $$ to show that in fact $d\phi d\xi =d\theta du$, and the identity follows.