I see there are two methods of u-substitution.
The first (and likely more common) is to set u to part of the equation and sub out your x and integrate, sub x back in and evaluate on the original [a,b].
There is a second method where you sub out x and re-evaluate a and b based on u (ie. x = 1 - > 0, u = 1 -> 1), then integrate and evaluate the integral based on this new [a,b].
Is there a difference in the application of these methods? Can I choose freely which one to use? Or is it based on the given problem?
There is a slight difference between the methods, but nothing big. Integration by substitution is essentially a change of coordinates. Under this way of thinking about the problem, integration by substitution is the process of taking the integral $$ \int_{A} f(x)\thinspace dx $$ with $F:A \rightarrow \mathbf{R}$, and precomposing with a function $g:B \rightarrow A$ so that the above integral becomes $$ \int_{g^{-1}(A)} f(g(x))\thinspace {dg \over dx}dx$$ You might notice that this looks a lot like the chain rule, as we are composing functions just like in the chain rule. Also notice the ${dg \over dx}$ before the infinitesimal length $dx$. This derivative is there because the function $g$ may distort length/area/etc., changing the value of the integral. Placing this derivative before the $dx$ corrects for such distortion.
Now for the difference: In the first method you give, what you are doing is taking the integral of $f(g(x))$ over $g^{-1}(A)$, and then applying $g$ again so that you evaluate $g$ of the integral of $f(g(x))$ over $g(g^{-1}(A))=A$. You will get the same result as the second method, but with the added step of applying $g$ to whatever expression you obtain for the integral. Perhaps this could make things simpler when $g^{-1}(A)$ is a complicated region while $A$ is not. Regardless, usually the second method you give is known as "integration by substitution". The first method is something like "integration by substitution and then unsubstitution".