Integrals by substitution: general case? I am trying to understand a general case for the substitution rule.
My real case is this:
$$\int{}(\sqrt{9-t^2})(-2t)dt$$
Making a generality.  If I have an integral in the form of:
$$\int{ab}$$
Where a need to be treated by the SUBSTITUTION RULE and b can be compute directly.
Procedure:
a. compute the integral for u
a=u
$\int{(u)}du=\frac{u^2}{2}du$
After that I need to make the substitution back.  But what happens whit the b?
Option 1. be remains the same (don't look for me like the correct answer)
$\frac{u^2}2(b)$
or,
Option 2. (I don't feel absolute comfortable with this)
$\frac{u^2}2(\frac{b^2}2)$
or,
Option 3.
None of the above If is the case please explain the correct path
 A: What you seem to be neglecting is accounting for $du$ when you permform substitution.
E.g.
$$\int{}\sqrt{\color{blue}{\bf(9-t^2)}}\underbrace{(-2t)dt}_{du}$$
Here, we can easily choose $\color{blue}{\bf u = 9 - t^2}$. But note, if we do this, we also need to substitute the equivalent of $-2t\,dt$, as well. We are fortunate with this integrand because it so happens that  $$\color{blue}{\bf u = 9 - t^2} \implies \color{red}{\bf du = -2t \,dt}$$ 
This gives us $$\int \sqrt{(\color{blue}{\bf 9 - t^2})} \color{red}{\bf (-2t \,dt)} = \int \color{blue}{\bf u}^{1/2} \color{red}{\bf du} = \dfrac {u^{3/2}}{3/2} + C = \dfrac 23u^{3/2} + C  $$ $$ = \frac 23 (9 - t^2)^{3/2} + C$$
Note that your integral is of the form $$\int \sqrt{9 - t^2}(9-t^2)'\,dt = \int g(f(t)) f'(t) \,dt$$ and our corresponding substitutions are $u = f(t), \; du = f'(t)\,dt$, giving us $$\int g(u) \,du$$ 
A: Suppose you know that $\,F'(x)=f(x)\,$ and that $\,g\,$ is a differentiable function, then
$$\int g'(x)\,f(g(x))\,dx=F(g(x))+C(=\text{a constant})$$
In your example, $\;-2t= \left(9-t^2\right)'\;$ , and since $\;\int \sqrt x\,dx=\frac23 x^{3/2}+C\;$ , we get
$$\int-2t\sqrt{9-t^2}dt=\frac23(9-t^2)^{3/2}+C$$
