What allows one to use a change of variables?

I find myself often using a change of variables or coordinates to solve various integration problems. Some are easy to justify for me such as integrating over a circle in Cartesian vs polar coordinates, yet some feel quite arbitrary, for instance $\int \sqrt{5+4x-x^2}$, Let $x-2 = 3\sin\theta$. From an analysis perspective, what lets me know that the things im trying to integrate are still really the same?

Does this fundamentally come down to the (in this example) statement that $dx = 3\cos\theta d\theta$?

thanks!

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The change of variable theorem is usually proved rigorously in Analysis books. Are you familiar with the definition of the Riemann integral and Darboux sums? It's pretty easy if you are. – Ayman Hourieh Feb 6 '13 at 21:23
yep i know both, but i dont think ive ever seen a rigorous proof. – MSEoris Feb 6 '13 at 21:26
Do you know about differential forms? – Neal Feb 7 '13 at 0:20
@Neal not really, is it some thing i can get with a quick reference? does it help to understand why we can change variables? – MSEoris Feb 7 '13 at 1:17
You don't need differential forms for change of variables, but it makes the statement very simple. – cats Feb 7 '13 at 5:50

In one dimension, the fundamental theorem of calculus proves the change of variables formula. If $f$ is continuous and $g \in \mathcal{C}^1,$ then $\displaystyle\int_{a}^{b} f(g(t))g^{\prime}(t) dt = \int_{g(a)}^{g(b)} f(x) dx.$ How do we prove this? Well, $f(g(t))g^{\prime}(t)$ is continuous by assumption, so it's integrable, so the left hand side exists; the right hand side obviously exists as well. Since $f$ is continuous, there exists $F$ such that $F^{\prime} = f,$ and $(F(g(t))^{\prime} = F^{\prime}(g(t))g^{\prime}(t) = f(g(t))g^{\prime}(t),$ from which LHS $= F(g(b))-F(g(a)) =$ RHS.

The general change of variables formula in higher dimensions is a little more complicated to prove, but it's just a lot of book-keeping.

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Completing the square, you have that

$$\sqrt{5+4x-x^2} = \sqrt{9-(x-2)^2}$$

and the trick is to recognize that the identity $\sin^2 t + \cos^2 t = 1$ helps you to simplify the integrand. The main goal is to get rid of the square root. If we had had $\sqrt{1-x^2}$ you could just have let $x=\sin \theta$ and the square root would simplify.

To handle your integrand you need to adjust a little: you want $9-(x-2)^2$ to simplify. By letting $(x-2)=3\sin \theta$, you get $9-(x-2)^2 = 9(1-\sin^2 \theta) = 9\cos^2 \theta$ and everything will work out beautifully.

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yes this is how you solve this example problem, but im more interested in fundamentally what allows us to make such changes. – MSEoris Feb 6 '13 at 20:38

If $y=\int\sqrt{5+4x-x^2}dx$, then $\frac{dy}{dx}=\sqrt{5+4x-x^2}$. If $(x-2)=3\sin\theta$,$\frac{dx}{d\theta}=3\cos\theta$. Now by the chain rule, we have

$$\frac{dy}{d\theta}=\frac{dy}{dx}\frac{dx}{d\theta}=\sqrt{5+4x-x^2}\times3\cos\theta=9\cos^2\theta$$

Now just integrate both sides with respect to $\theta$ to solve for $y$. Does this answer your question?

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