Let $f$ be a cont. on $\mathbb{R}$ and define $G(x)=\int_0^{\sin (x)}f(t) dt $. Show that $G$ is differentiable on $\Bbb{R}$ and compute $G'$. 
Let $f$ be a continuous function on $\mathbb{R}$ and define
$$G(x)=\int_0^{\sin (x)}f(t) dt $$ Show that $G$ is differentiable on
$\mathbb{R}$ and compute $G'$.

This is an exercise from Elementary Analysis: The Theory of Calculus by Kenneth A. Ross.
My first idea is the following. Let $x\in \Bbb R$. Define $F(x):=\displaystyle\int_0^{x}f(t) dt $. And prove that $G:=F\circ \sin$ is differentiable at $x$ using:

It is clear that $\sin(x)$ is differentiable at $x$. So I need to proof that $F$ is differentiable at $y:=\sin(x)$. I was trying to prove that using:

34.3 Fundamental Theorem of Calculus II.
Let $f$ be an integrable function on $[a, b] .$ For $x$ in $[a, b],$ let
$$
F(x)=\int_{a}^{x} f(t) d t
$$
Then $F$ is continuous on $[a, b] .$ If $f$ is continuous at $x_{0}$ in $(a, b),$ then $F$ is differentiable at $x_{0}$ and
$$
F^{\prime}\left(x_{0}\right)=f\left(x_{0}\right)
$$

I'm not sure if I can use this theorem. Obviously, $f$ sattisfies the conditions. But as $y=\sin(x) \in [-1,1]$, I dont see how I could choose $a,b$ in theorem 34.3, such that it fits this case. It seems clear that I should choose $a:=0$, and if $y>0$ I could choose $b:=1$, but if $y<0$, I can't find any $b$ such that $y\in[a,b]$.

I wanted to know if I'm heading in the right direction, so I looked at the official solution manual. What they are doing there seems complete nonsense to me:

I don't see why they are doing what they are doing. They end the proof with $G$ is continuous. Why is that result needed ? And isn't $f$ confused for (some undefined) $F$ in the beginning?

As this proof didn't satisfy me, I looked further, and I found this proof:

This proof seems completely solid to me. But it is much less intuitive (for me). I wouldn't have come with this proof myself.
My question are:

*

*Was I heading in the right direction with my first idea ? Or is it not possible to apply theorem 34.3 in this kind of way ?

*Am I right that the official solution manual is complete nonsense, or am I missing something ?

*Is this last proof I found correct, and do you think the author of the book has this kind of proof in mind when he wrote this exercise ? Or do you think the author was expecting some other kind of proof ?

Edit: For the bounty I would like to see a rigorous prove of why $G$ is differentiable on all $x\in \Bbb R$ using only theorems that are proven in Elementary Analysis: The Theory of Calculus by Kenneth A. Ross.
 A: We have
$$
\begin{align}
G(x)
&=F(\sin(x))\\
&=\int_0^{\sin(x)}f(t)\,\mathrm{d}t\tag{1}
\end{align}
$$
where
$$
\begin{align}
F(u)
&=\int_0^uf(t)\,\mathrm{d}t\\
&=\underbrace{\int_{-1}^uf(t)\,\mathrm{d}t}_{\substack{\text{differentiable}\\\text{for $u\ge-1$}}}-\underbrace{\int_{-1}^0f(t)\,\mathrm{d}t}_\text{constant}\tag{2}
\end{align}
$$
The Fundamental Theorem of Calculus says
$$
F'(u)=f(u)\tag{3}
$$
The Chain Rule says that
$$
\begin{align}
G'(x)
&=F'(\sin(x))\cos(x)\\
&=f(\sin(x))\cos(x)\tag{4}
\end{align}
$$
Technical Point: To apply Theorem 28.4 to Theorem 34.3 without any further work, we should choose the lower bound of integration in $(2)$ to be lower than $-1$.
However, with a bit of extra work, we can show that $f$ need only be continuous on $[-1,1]$. The only problem arises when computing $F'(\pm1)$. Since $F(u)$ will only see $u\in[-1,1]$, we only need to consider the one-sided derivative at $\pm1$.
Suppose $u_0=\pm1$, then for $u\in(-1,1)$ we can apply the Mean Value Theorem to find a $\xi$ between $u_0$ and $u$, hence in $(-1,1)$, so that
$$
\begin{align}
\frac{F(u)-F(u_0)}{u-u_0}
&=F'(\xi)\\
&=f(\xi)\tag{5}
\end{align}
$$
Therefore, since $f$ is continuous on $[-1,1]$, we have the one-sided derivative
$$
\begin{align}
F'(u_0)
&=\lim_{u\to u_0}\frac{F(u)-F(u_0)}{u-u_0}\\
&=\lim_{\xi\to u_0}f(\xi)\\
&=f(u_0)\tag{6}
\end{align}
$$



*

*You are thinking correctly about applying the Fundamental Theorem of Calculus and the Chain Rule.

*I am not sure why they are showing that $G$ is continuous. Since they have shown its derivative exists, it is already continuous.

*The handwritten proof looks okay. It is a different approach, but it is valid. You don't need the Chain Rule with that approach.
