Why do we use substitutions when computing integrals? I was reading through Spivak's Calculus, and in the last chapter of integrals he explains how 

$$\int_{a}^{b} (f \circ g)(x) \cdot g'(x) \, \mathrm{d}x = \int_{g(a)}^{g(b)} f(x) \, \mathrm{d}x. $$ 

He discusses later a lot of ways to compute integrals $g(x)=u$, $ g'(x)= du $ etc..  
My question is, why is all of this necessary? Why not just rely on the first one?  
I've done all the ones he used substitution in, only relying on the original theorem, and I have found it confusing. Browsing through MSE, people substitute instead of just trying to show $(f\circ g)(x) \cdot g'(x)$  
Is learning substitution essential for computing advanced integrals?
 A: The whole u = g(x) stuff is essentially a way of working out how to write it in that form - if you can just do it by recognition (a skill that usually comes through doing lots by substitution) then that's great.
A: Matt Rigby has the right of it, for the most part.  However, the other thing to think about is this: integrals can be very complicated.  Often, they will require more than one simplification to handle; so, perhaps you do a substitution first, but then will need to do integration by parts or another substitution or a trig sub on the result; etc.  Maybe you can see your way through an integral that takes five different solution steps... but most people can't!  
The substitution theorem allows us to rewrite in terms of a nicer integral, and never look back.
A: The expression you started with proves that a properly computed substitution is equal to the original integral.  That is why doing it following exactly your basic expression always works. If you look at your expression, you can see that substitution is the chain rule.  The way people do it is a shortcut, rather than writing out every detail. 
How you go about evaluating integrals is to some extent a matter of taste -- there are often equivalent ways of doing things, and you may find you have a preference for one rather than another.  However, you might as well simplify things as much as you can.  As Nicholas points out, integrals can get very complicated.  If you hit on a substitution that makes the problem easier, go for it.
Of course there are far more definite integrals that cannot be evaluated by elementary methods than those that can.  For practical purposes, if you have a real mess, a numeric method is probably the best approach; but if you are taking a class, get that substitution working.
A: In your original equation:
$$\int_{a}^{b} (f \circ g)(x) \cdot g'(x) \, dx = \int_{g(a)}^{g(b)} f(x) \, dx$$ 
Notice that:
$$\int_{a}^{b} (f \circ g)(x) \cdot g'(x) \, dx = \int_{a}^{b} f(g(x))  \, dg(x)$$
And so:
$$\int_{a}^{b} f(g(x))  \, dg(x) = \int_{g(a)}^{g(b)} f(x) \, dx$$
In other words, if you integrate along the density $g(x)$ between $a$ and $b$, it is the same as integrating along the uniform density $x$ between $g(a)$ and $g(b)$.  So integration is parameter independent, and finding the right density $g$ could be greatly simplifying.
