Let $f$ be a continuous function on $\mathbb R$ and define $$G(x) = \int _0 ^{\sin x}f(t)dt$$ for $x \in \mathbb R$. Show that $G$ is differentiable on $\mathbb R$ and calculate its derivative.
Attempt: Let $h >0$. Then, $$\frac {G(x+h) - G(x)}{h} = \frac {\int_{\sin(x)}^{\sin(x+h)}f(t) dt}{h}.$$
Let $c,t \in \mathbb R$. Then, $\forall\varepsilon, \exists\delta>0$ s.t. if $|t-c|<\delta, |f(t)-f(c)|<\varepsilon.$ This implies that $f(c) - \varepsilon < f(t) < f(c) + \varepsilon.$
Take integral on each side and divide by $h$. Then, we have $$\frac{\int_{\sin (x)}^{\sin(x+h)}(f(c)-\varepsilon) dt}{h} \le \frac{\int_{\sin (x)}^{\sin(x+h)}f(t)dt}{h} \le \frac{\int_{\sin (x)}^{\sin(x+h)}(f(c)+\varepsilon)dt}{h}.$$
This implies : $(\sin(x+h)-\sin(x))(f(c) -\varepsilon)/h \le \frac{\int_{\sin (x)}^{\sin(x+h)}f(t)dt}{h}\le (\sin(x+h)-\sin(x))(f(c) +\varepsilon)/h.$
I am stuck here. I was planning to show $|\frac{\int_{\sin (x)}^{\sin(x+h)}f(t)dt}{h}- \cdot| < \varepsilon.$ But, I don't know what to do here. Could you give some help?
If $F(x)=\int_0^xf(t)\,\mathrm dt$, then $G(x)=F\bigl(\sin(x)\bigr)$. Now, use the chain rule and the fundamental theorem of Calculus.