Substitution of an implicit variable I wasn't sure how to title this question: I want to manipulate the integral
$$I(a,b) = \int_0^{\frac{\pi}{2}} \frac{d \phi}{\sqrt{a^2\cos^2 \phi + b^2 \sin^2 \phi}}$$
with this subsitution: $$\sin \phi = \frac{2a \sin \psi}{a+b+(a-b)\sin^2 \psi}$$
I have enough hints for the later calculations, but I have a problem starting: How do I express $d\phi$ with $d \psi$? All substitutions I ever did were of the form $x = g(t)$ but never $\sin(x) = g(t)$. How does the differentiation work? I think the problem is that I don't really understand the meaning of the $d$ in front of a term. What kind of object is $d \phi$ and how and why can I calculate with it, like $\frac{d \phi}{d \psi}$?
 A: What you formally do is "differentiate on both sides". This is also called implicit differentiation.
Thus the equation $\sin(x)=g(t)$ implicitly differentiated becomes
$$\cos(x) dx=g^\prime(t) dt\tag{1}$$
This makes a lot of sense when you imagine $t=t(x)$ as a function of $x$ and simply derive both sides by the independent variable $x$ using the chain rule:
$$\frac{d}{dx}(\sin(x))=\frac{d}{dx}(g(t(x)))$$
This yields
$$\cos(x)=g^\prime(t(x)) \frac{dt}{dx}\tag{2}$$
This is the rigorous interpretation of equation $(1)$.
As you have probably noticed, on this level, the differentials $dt$, $dx$ and 
so forth are nothing but useful notation with no intrinsic meaning.
Addendum:
Since you asked: there is a way to give intrinsic meaning to these differentials. Namely, in the theory of differential forms.
In this context, also the differential map $d$ is given a precise, rigorous meaning. Look it up in a book on Differential Geometry/Global Analysis. One which I like is


*

*H. Amann, J. Escher, Analysis III.


In Germany this is sort of a standard text book, but it also has an excellent English version.
A: Hint
$$\sin \phi = \frac{2a \sin \psi}{a+b+(a-b)\sin^2 \psi}$$ gives $$\phi=\sin ^{-1}\left(\frac{2 a \sin (\psi )}{a+b+(a-b) \sin ^2(\psi )}\right)=\sin ^{-1}\Big(u(\psi )\Big)$$ Now, apply the chain rule to get $$d\phi=\frac{u'(\psi )}{\sqrt{1-u(\psi )^2}} d\psi$$
