Quick question about an article about Watson triple integral

Question

So, I've recently read about Watson's triple integral, in particular i was interested in the first one. All this comes from: http://mathworld.wolfram.com/WatsonsTripleIntegrals.html. Now I've got through the derivation for $$I_1=\int_0^\pi\int_0^\pi\int_0^\pi\frac{dudvdw}{1-\cos{u}\cos{v}\cos{w}}$$ (I will be ignoring the $\pi^3$ term). Referencing to the link, I'm quite unsure about the step from (23) to (24). We have that: $$4\int_0^{\pi/2}\int_0^{\pi/2}\int_0^\infty\frac{\sin{\theta}drd\theta d\phi}{1+r^4\sin^4{\theta}\cos^2{\theta}\sin^2{\phi}\cos^2{\phi}}=$$ $$=4\int_0^{\pi/2}\int_0^{\pi/2}\int_0^\infty\frac{\sin{\theta}drd\theta d\phi}{1+\frac{1}{4}r^4\sin^4{\theta}\cos^2{\theta}\sin^2{2\phi}}=$$ The OP says: we write $\psi=2\phi$ and continues: $$=4\int_0^{\pi/2}\int_0^{\pi/2}\int_0^\infty\frac{\sin{\theta}drd\theta d\psi}{1+\frac{1}{4}r^4\sin^4{\theta}\cos^2{\theta}\sin^2{\psi}}$$ But shouldn't also $d\psi=2d\phi$ and the bounds of the first integral change to $0,\pi$? The result of this integral, given in the link, which is $\frac{1}{4}\Gamma^4(\frac{1}{4})$ should be correct according to more articles. What am i missing?