Fundamental Theorem of Calculus, application I want to derive the function $$F(x)=\int_a^{x^2}\sin^3t\,dt$$ with the fundamental theorem of calculus, but I dont know how to handle the $x^2$.
Maybe with subsitution I think 
Fundamental theorem of calculus: Let $f:[a,b]\to\mathbb{R}$ a continuous function. Then the function $$F:[a,b]\to\mathbb{R},\; x\mapsto \int_a^xf(t)dt$$ is continuous in $[a,b]$ and differentiable in $(a,b)$ and $\frac{\partial F(x)}{dx}=f(x)$.
Could you help me to apply the theorem?
I only know how to do it without the theorem, if you use $$\sin^3(t)=\frac{d(\frac{1}{3}\cos^3t-\cos t))}{dt}$$ 
 A: Hint: Use the Chain Rule $$f(g(x))' = f'(g(x)) g'(x)$$
where $$f(x) = \int_a^{x} \sin^3 t \,\,dt$$
and $g(x) = x^2.$
A: one form of the fundamental theorem of calculus is $$d\left(\int_a^b f(t) \, dt \right) = f(b)\, db - f(a) \, da \tag 1$$  if you keep the lower limit fixed at $a$ and the upper limit is a variable $x^2,$ then $(1)$ becomes 
$$d\left(\int_a^{x^2} \sin^3 t\, dt \right) = \sin^3\left(x^2\right)\, d\left(x^2\right) - \sin^3(a) \times 0 =  2x\sin^3\left(x^2\right)\, dx$$  now, dividing by $dx$  gives $$\frac{d}{dx}\left(\int_a^{x^2} \sin^3 t\, dt \right) = 2x\sin^3\left(x^2\right).$$
A: $\displaystyle F(x)=\int_a^{x^2}\sin^3t\,dt$
Let $\displaystyle f(x) = \int_a^x\sin^3t\,dt$
Note that the variable $t$ is unimportant. It would mean exactly the same thing if we had said
Let $\displaystyle f(x) = \int_a^x\sin^3u\,du$
We know that $f'(x) = \sin^3x$
So $f'(x^2) = \sin^3(x^2)$.
Next, let $g(x) = x^2$.
We know that $g'(x) = 2x$.
Then $\displaystyle f(g(x)) = \int_a^{x^2}\sin^3t\,dt = F(x)$.
The chain rule tells us that
\begin{align}
  F'(x)
  &= f'(g(x)) \cdot g'(x) \\
  &= f'(x^2) \cdot g'(x) \\
  &= \sin^3(x^2) \cdot(2x) \\
  &= 2x \sin^3(x^2)
\end{align}
