Is this true about integrating composite functions? Let's say that I'm integrating a composite function, say $f(g(x))$, that is in a form to which I can apply the substitution rule.  Is it true to say that both $f$ and $g$ must be differentiable?
I understand that the substitution rule requires $g$ to be differentiable and that the substitution rule relies on the chain rule, and the chain rule requires both f and g to be differentiable.
 A: 

Substitution rule relies on chain rule and the chain rule requires both f and g to be differentiable.


In showing $\displaystyle \int f(x)dx=\displaystyle \int f(g(t))g'(t)dt$,
All we require in substitution rule is that $f(x)$ be continuous so that its antiderivative exists-and is differentiable, say $F$ and $g(x)$ to have a continuous derivative, so the composition $F \circ g$ is differentiable, which we differentiate using chain rule.
So, by chain rule, $\displaystyle\frac{d}{dt}(F \circ g)(t)=f(g(t))g'(t)$. By fundamental theorem of calculus,
$\displaystyle \int_a^b f(g(t))g'(t)dt=(F \circ g)(b)-(F \circ g)(a)=\displaystyle \int_{g(a)}^{g(b)}f(x)dx$. Thus we have proved substitution rule.
Since we don't need to differentiate $f \circ g$ to prove substitution rule, we don't require $f$ to be differentiable.
A: if you talk about the Riemann-integral: if $f : [a,b] \to [c,d]$ R-integrable and $g : [c,d] \to \mathbb{R}$ continuous, than $g \circ f : [a,b] \to \mathbb R$ is R-integrable.
Differentiable implies continuous, so if f,g are differentiable they are R-integrable
