Why is it not true that $\int_0^{\pi} \sin(x)\; dx = 0$? I know the following is not right, but what is the problem. So we want to calculate 
$$
\int_0^{\pi} \sin(x) \; dx
$$
If one does a substitution $u = \sin(x)$, then one gets
$$
\int_{\sin(0) = 0}^{\sin(\pi) = 0} \text{something}\; du = 0.
$$
We know that $\int_a^a f(x) \; dx = 0$ for all functions $x$, so why doesn't this work for the above?
I get that the "something" "can't be found" because $du = \cos(x)\,dx$. But does it really matter what the $du$ is when one is integrating from $0$ to $0$?
Edit: I don't know what a "diffeomorphism" is. I am just in basic calculus.

Very precise answers have been given. I would like to answer the subject question only "Why is it not true that $\int_0^{\pi} \sin(x)\; dx = 0$?" 
This is first a matter of visual evidence and common sense. 
In the interval $[0,\pi]$, your sine function is positive, and stricly inside $]0,\pi[$. With a little calculus, you obtain that $\sin$ is stricly positive (say, above $\sin(a)$) in some $[a,\pi-a]$ interval, $0<a<\pi-a<\pi$. Thus your integral, which can be interpreted as an area here, should be above $\sin(a)\times (\pi - 2a)>0$. 

Based on these observations, you should resign to the fact that your substitution is not valid for some reason (explained in the above precise answers). Your trick reminds me of the "proof" that $1=0$, which does unappropriate substitutions, leading to a wrong conclusion.
Logic and observations are good masters in maths.
