How to calculate the derivative of this integral? Here it is :
$$
\frac{\mathrm d}{\mathrm dx}\left( \int_{\cos x}^{\sin x}{\sin \left( t^3 \right)\mathrm dt} \right)
$$
I've got the answer but I don't know how to start , what to do ?
Here is the answer : 
$
\sin \left( \sin^3 x \right)\cos x + \sin \left( \cos ^{3}x \right)\sin x
$
So first I calculate the primitive and then I derivate it. But I don't know how to integrate. Should I use 'substitution' method ? I tried but then i was blocked...
 A: Look up Leibniz integral rule
$$\frac{d}{dx}\int_{a(x)}^{b(x)} f(t,x)\,dt = \frac{d b(x)}{d x}\,f(b(x),x)-\frac{d a(x)}{d x}\,f(a(x),x)+ \int_{a(x)}^{b(x)}\frac{\partial}{\partial x}\,f(t,x)\,dt$$
A: First put the integrate as 
$\int_0^{\sin x} \sin(t^3)\mathrm dt - \int_0^{\cos x} \sin(t^3)\mathrm dt$
Then derivate the two items separately using the formula for the derivative of an integral with a varying upper integrating bound, e.g.,
$$\frac{\mathrm d}{\mathrm dx} \int_0^{\sin x} \sin(t^3)\mathrm dt = \sin((\sin x)^3)(\sin x)' = \sin((\sin x)^3) \cos x.$$ 
Hope this can help you.
A: I understand from the comments that you are not completely pleased with the answers so far. That's why I try it (with a bit delay). Note that there is nothing new in this answer ...
All you need to know is the fundamental theorem of calculus
$$f(x) = \frac{d}{dx} F(x)$$
with
$$F(x) = \int^x_a f(t) dt$$
and the chain rule
$$\frac{d}{dx} f[g(x)] = f'[g(x)] g'(x).$$
Your integral is given by
$$ \int_{\cos x}^{\sin x}{\sin ( t^3) \,dt} =F(\sin x) - F(\cos x)$$
with $$F(x) = \int_a^x f(t) dt$$
and $f(t)=\sin(t^3)$.
Therefore,
$$ \frac{d}{dx}\left[ \int_{\cos x}^{\sin x}{\sin ( t^3 ) dt} \right]
 = \frac{d}{dx} [F(\sin x) - F(\cos x)]
 = F'(\sin x) \sin' x - F'(\cos x) \cos' x$$
$$ = f(\sin x) \cos x + f(\cos x) \sin x = \sin ( \sin^3 x) \cos x + \sin (\cos^3 x) \sin x.$$ 
