How to calculate the derivative of this integral?

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, $$ \frac{d}{dx}\left[ \int_{\cos x}^{\sin x}{\sin ( t^3 ) dt} \right] = \frac{d}{dx} [F(\sin x) - F(\cos x)] = F'(\sin x) \sin' x - F'(\cos x) \cos' x$$ $$ = f(\sin x) \cos x + f(\cos x) \sin x = \sin ( \sin^3 x) \cos x + \sin (\cos^3 x) \sin x.$$

Look up Leibniz integral rule

$$\frac{d}{dx}\int_{a(x)}^{b(x)} f(t,x)\,dt = \frac{d b(x)}{d x}\,f(b(x),x)-\frac{d a(x)}{d x}\,f(a(x),x)+ \int_{a(x)}^{b(x)}\frac{\partial}{\partial x}\,f(t,x)\,dt$$

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.