Substitution for definite integrals In my experience, Calculus II students dislike changing bounds in definite integrals involving substitution.  When facing an integral like
$$\int_0^{\sqrt{\pi }} x \sin \left(x^2\right)dx,$$
for example, most US Calc II students would introduce $u=x^2$ and compute
\begin{align}
  \int x \sin \left(x^2\right)dx &= \frac{1}{2}\int \sin(u) \, du \\
  &= -\frac{1}{2} \cos(u)+c = -\frac{1}{2} \cos(x^2)+c.
\end{align}
Afterward, they would conclude that
$$\int_0^{\sqrt{\pi }} x \sin \left(x^2\right) \, dx = -\frac{1}{2} \cos(x^2) \big|_0^{\sqrt{\pi}} = 1.$$
I would generally encourage them to write
\begin{align}
  \int_0^{\sqrt{\pi }} x \sin \left(x^2\right)dx &= \frac{1}{2}\int_0^{\pi} \sin(u) \, du \\
  &= -\frac{1}{2} \cos(u) \big|_0^{\pi} = 1.
\end{align}
This question expresses opinion of a typical such student and this answer correctly expresses the fact that the two step process favored my most calculus students is actually more work.
I think there's more to it than this, though.  Specifically, the identity
$$\int_a^b f(g(x)) g'(x) \, dx = \int_{g(a)}^{g(b)} f(u) \, du$$
is a relationship between definite integrals which could have applications other than symbolic evaluation of the integral on the left.  In this case, the change of the bounds of integration is important in its own right.  Thus my question:
What are some important applications of change of variables in definite integration, other than symbolic evaluation?
I have at least one answer but would be happy to hear more, particularly those that are easily understandable by Calc II students, as I think it's an important pedagogical question.
 A: 
What are some important applications of change of variables in definite integration, other than symbolic evaluation?

Here are a handful:


*

*To give a computational (as opposed to geometric) proof that
$$
\int_{-a}^{a} f(x)\, dx
  = \begin{cases}
    0 & \text{if $f$ is odd}; \\
    2\displaystyle\int_{0}^{a} f(x)\, dx & \text{if $f$ is even.}
  \end{cases}
$$

*To show that if $a$ and $b$ are positive real numbers, then
$$
\int_{a}^{ab} \frac{1}{t}\, dt = \int_{1}^{b} \frac{1}{t}\, dt,
$$
which, of course, is the key to proving $\log(ab) = \log(a) + \log(b)$.

*To prove that if $f$ is continuous on $[-1, 1]$, then
$$
\int_{0}^{2\pi} f(\cos \theta) \sin\theta\, d\theta
  = \int_{0}^{2\pi} f(\sin \theta) \cos\theta\, d\theta
  = 0.
$$
A: Here's another common example:
$$ \int_a^b f(x)\,dx = \int_a^b f(a+b-x)\,dx.$$
The geometric interpretation is simply that the area of a region is preserved by reflection about its midsection.

A: I will give my opinion as a student who recently "finished" Calculus, i.e. from Calc. I to Real Analysis.
First of all, I'd point out that the "symbolic evaluation" is probably the most important part of it. The identity that you've written was, for me, the only way I could actually accept the otherwise cryptic statement, presented to us as a "rule" for indefinite integrals: $$\mathrm du = u'(x)\mathrm dx$$
How was I supposed to reason this? What is this mysterious $\mathrm d$? A function of functions, whose result ends up having another such $\mathrm d$? An infinitesimal?  What's that?
"Intuition" was never a deal-sealer for me, and under no circumstances was I prepared to accept "multiplying by $\mathrm dx$" in the identity $\mathrm du/\mathrm dx = u'(x)$. I needed to know why. This was where the identity for definite integrals came in. Of course, knowledge of a formal proof, one can argue, doesn't necessarily constitute an understanding; the "why". However, what it does provide is a sort of tether, a security that one can start with the mathematics that they (for whatever reason) know and accept at that moment, and arrive if they wish at the conclusion. So, once I knew and understood the proof concerning areas under functions, the proof that $$\int_a^b f(g(x)) g'(x) \,\mathrm dx = \int_{g(a)}^{g(b)} f(u) \,\mathrm du$$
I was able to keep on considering the mysterious $\mathrm d$ as simple notation for indicating which is the independent variable, and as for indefinite integrals, I could justify the "cryptic statement" I mentioned by considering the definite integral with a variable limit.
In summary, the rule for changing variables in definite integrals, and the proof of said rule with reasoning that was within my grasp at the time, served as psychological grounding when dealing with $\mathrm dx$ and the likes. Note that this could be extended to (a limited set of) other contexts, notably the "separation of variables" one realizes in certain differential equations. Without this I was supposed to believe, again, that I was actually multiplying by $\mathrm dx$!
I realize that you ask for applications other than symbolic evaluation, and though I've mentioned the separation of variables in ODEs, which I think is a good example, I'll try to think of some more and edit this post later.
A: Substitution in definite integration is the basis for understanding how a general normal distribution is related to the standard normal distribution.  This is a very important application.
The formula for the general, normal distribution with mean $\mu$ and standard deviation $\sigma$ is
$$f_{\mu,\sigma}(x) = \frac{1}{\sqrt{2\pi}\sigma}e^{-(x-\mu)^2/(2\sigma^2)}.$$
The standard normal distribution is obtained by setting $\mu=0$ and $\sigma=1$. Thus,
$$f(x) = f_{0,1}(x) = \frac{1}{\sqrt{2\pi}}e^{-x^2/2}.$$
Each of these is an example of a probability distribution, i.e. a non-negative, integrable function $f:{\mathbb R}\rightarrow\mathbb R$ whose overall integral is $1$.  If $X$ is a random variable with probability distribution $f$, then we compute the probability that $X$ lies in some interval $[a,b]$ via
$$P(a\leq X\leq b) = \int_a^b f(x)\,dx.$$
Now, the normal distribution is very ubiquitous occurring in reality with amazing frequency.  This is the essential content of the central limit theorem.  Thus, computing integrals of the form
$$\frac{1}{\sqrt{2\pi}\sigma} \int_a^b e^{-(x-\mu)^2/(2\sigma^2)}dx$$
is a very important problem.  Unfortunately, there is no simple closed form expression for the anti-derivative.  The standard way around this problem, as taught in elementary statistics, is to translate the computation to the standard normal.  As there is just one standard normal, we can look up values for $P(0\leq X\leq b)$ in a pre-computed table.  The translation states, if $X_{\mu,\sigma}$ is normally distributed with mean $\mu$ and standard deviation $\sigma$, then 
$$P(a\leq X_{\mu,\sigma}\leq b) = P\left(\left(\frac{a-\mu}{\sigma}\right) \leq X_{0,1} \leq P\left(\frac{b-\mu}{\sigma}\right)\right).$$
The mystery is, where do those numbers $(a-\mu)/\sigma$ and $(b-\mu)/\sigma$ come from? And the answer is, from changing bounds in a definite integral involving substitution!  Indeed, letting $u=(x-\mu)/\sigma$ so that $\frac{1}{\sigma}dx=du$, we get
\begin{align}
  \frac{1}{\sqrt{2\pi}\sigma} \int_a^b e^{-(x-\mu)^2/(2\sigma^2)}dx &=
  \frac{1}{\sqrt{2\pi}} \int_a^b e^{-\frac{1}{2}\left((x-\mu)/\sigma\right)^2}
    \frac{1}{\sigma}dx \\
  &= \frac{1}{\sqrt{2\pi}} \int_{(a-\mu)/\sigma}^{(b-\mu)/\sigma} e^{-\frac{1}{2}u^2} du.
\end{align}
Note that intuitive rules in basic statistics, like 70% of a population is within one standard deviation from the mean or 95% within two standard deviations, follows from this.  The 95%, for example, comes from the result of the following numerical computation:
$$\frac{1}{\sqrt{2\pi}} \int_{-2}^2 e^{-x^2/2} dx = 0.9545.$$
Speaking of numerical integration, that forms the basis for another answer...
A: Here's another interesting example:
We can use the substitution $u=x^k$ to show that
$$\int_0^{\infty} \frac{\sin(x^k)}{x}\,dx =
  \frac{1}{k}\int_0^{\infty} \frac{\sin(u)}{u}\,du.$$
Thus, the known result that
$$\int_0^{\infty} \frac{\sin(x)}{x}\,dx = \frac{\pi}{2},$$
yields
$$\int_0^{\infty} \frac{\sin(x^k)}{x}\,dx = \frac{\pi}{2k}.$$
