Is there ever a requirement to change the limits of integration? I don't have issues with doing integration problems, but occasionally I see the solution changing the limits of integration whenever a $u$-substitution is done.
I obviously don't have a problem doing this, and I just recently noticed my book doing this under the chapter involving "area of surface of revolution." 
My question is did I develop a bad habit by never changing the limits of integration, or is it a best practice to always change limits of integration?  
My teacher said on a test, and in general, if we do not change the limits of integration then we should be signifying this by labeling our limits of integration $x=$ lower-limit and $x=$upper-limit.
EDIT EXAMPLE INCLUDED
After further investigation, my confusion is because of the two below equations:
Find the exact area of the surface obtained by rotating the curve about the x-axis
$$y=\sqrt{1+4x}, 1\le x\le 5$$
The limits of integration were changed in the solution to this problem.  
The given curve is rotated about the y-axis.  Find the area of the resulting surface.
$$y=x^\frac{1}{3}, 1\le y \le 2$$
The limits of integration were NOT changed in the solution to this problem.  
 A: Sometimes we are doing an indefinite integral by $u$-substitution, and then the form of integral involving $u$ is easily converted back into $x$ or whatever the original variable was at the end of the process.
However when a definite integral is involved, you have a choice of either converting the limits of integration from (say) $x$ limits to $u$ limits, or considering the $u$-substitution as a means to obtaining the final indefinite integral in terms of $x$ and using the original limits of integration.
The former has the advantage of skipping the substitution back into $x$, but at the cost of figuring out how to change the limits of integration into terms of $u$.  This was a bad habit you learned, or more precisely, a useful habit you failed to learn.  Converting the limits from $x$ to $u$ is ordinarily just a matter of using the $x$ limits in the expression for $u$ in terms of $x$.
A: The idea is that the limits of the integral are in terms of the variable of integration. So take for instance:
$$
\int_{0}^{2\pi}\cos(x)e^{\cos x}dx
$$
Notice that our function is periodic with period $2\pi$ and symmetric about its half period point, $\pi$; therefore,
$$
\int_{0}^{2\pi}\cos(x)e^{\cos x}dx=2\int_{0}^{\pi}\cos(x)e^{\cos x}dx
$$
This will be important when we make the $u$-substitution. In this case, we might choose $u=\cos(x)$ with $du=-\sin(x)dx$; therefore,
$$
\int_{0}^{2\pi}\cos(x)e^{\cos x}dx\\=2\int_{0}^{\pi}\cos(x)e^{\cos x}dx=-2\int_{1}^{-1}\frac{e^{u}u}{\sqrt{1-u^2}}du=2\int_{-1}^{1}\frac{e^{u}u}{\sqrt{1-u^2}}du
$$
You'll notice a few things about this process. First, if I had left the limits of integration as $0$ to $\pi$ after making the $u$-substitution, then the two integrals would have been different since the functions being integrated over would have been different.  Remember, the integration variable is a dummy variable, so what you use for it is completely irrelevant. Thus,
$$
\int_{0}^{2\pi}\cos(x)e^{\cos x}dx=\int_{0}^{2\pi}\cos(u)e^{\cos u}du
$$
Second, the way you decide how the limits change is by evaluating the parameterization equation. So for the above example $u(x)=\cos(x)$ so $u(0)=1$ and $u(\pi)=-1$. If I had left the limits as $0$ and $2\pi$, then the $u$-integral would have been from $1$ to $1$ which is obviously wrong since it evaluates to $0$ and the graph of our function clearly has a net positive area over the interval.
Edit: As Barry Cipra alludes to in the comments, there is a neat explanation for why the integral from $0$ to $2\pi$ causes problems when the variable changes. Specifically, I used the following identities in the above calculations
$$
\cos(x)=u\\
\sin(x)=\sqrt{1-\cos^2(x)}\\
\therefore \ -\sin(x)=-\sqrt{1-u^2}
$$
But this is not always true. The second line comes from $\sin^2(x)+\cos^2(x)=1$, which means that you have to take a square root. And whenever a square root is introduced so too must a plus-or-minus sign. In other words, the correct identity should be
$$
-\sin(x)=\mp\sqrt{1-u^2}
$$
with the sign chosen based on the value of the input, $x$.
In this problem, $\sin(x)$ is positive from $0$ to $\pi$, but it is negative from $\pi$ to $2\pi$. So the result above could also be derived as follows:
$$
\int_{0}^{2\pi}\cos(x)e^{\cos x}dx=-\int_{1}^{-1}\frac{e^{u}u}{\sqrt{1-u^2}}du+\int_{-1}^{1}\frac{e^{u}u}{\sqrt{1-u^2}}du=2\int_{-1}^{1}\frac{e^{u}u}{\sqrt{1-u^2}}
$$
A: Suppose you have
$$
\int_0^{\pi/4} \left(\tan x\right)^3 \Big(\sec^2 x\,dx\Big)
$$
(where you should take the $\Big($Big parentheses$\Big)$  as a hint about what $u$ should be), then you can write
\begin{align}
u & = \tan x \\
du & = \sec^2 x\,dx
\end{align}
and then the integral is
$$
\int_{x=0}^{x=\pi/4} u^3 \,du = \left[\frac{u^4}4\right]_{x=0}^{x=\pi/4} = \left[\frac{(\tan x)^4}{4}\right]_0^{\pi/4} = \frac14.
$$
In this case, you could say changing the bounds is optional, although it speeds things up and simplifies things:
$$
\int_0^1 u^3\,du = \left[ \frac{u^4}{4}\right]_0^1 = \frac14.
$$
But there are situations in which you'll get nowhere if you don't change the bounds.  For example, the Beta function is defined by
$$
B(\alpha,\beta) = \int_0^1 x^{\alpha-1} (1-x)^{\beta-1}\,dx.
$$
So suppose we have
$$
\int_0^5 x^{\alpha-1}(5-x)^{\beta-1} \,dx.
$$
(I've seen integrals like this arising in statistics problems.)  Then let
\begin{align}
u & = x/5, \\
du & = dx/5.
\end{align}
The integral then becomes
$$
\int_0^1 (5u)^{\alpha-1} (5-5u)^{\beta-1}\,\Big(5\,du\Big) = 5^3 \int_0^1 u^{\alpha-1}(1-u)^{\beta-1}\, du = 125 B(\alpha,\beta). 
$$
Without changing the bounds, how could one recognize this as being the Beta function?
