Determine the second derivative of this integral at $\pi$ I have this question here which says:
Find $g''(\pi)$ where $g(y)=\int_{3}^{y^{2}} (\int_{0}^{\sin(x)} \sqrt{1+t^{2}} dt) dx$ 
Now I'm not really sure how to approach this. My intuition was to notice that the integral on the inside has a bound of $\sin(x)$ so I figured it must be a constant which I could pull out.
I could then use the fundamental theorem of calculus in order to take the derivative of the integral that's left over and then take the derivative again and solve that. I feel like that's wrong though since I have an integral left over.
My next idea was to just use the fundamental theorem of calculus normally so I would take the derivative of the outermost integral using the fundamental theorem of calculus.
I would then substitute a $y^2$ for the $t^2$ and multiply by the derivative of $t^2$ or just $2t$ and I would end up with:
$\int_{0}^{\sin(x)} 2y\sqrt{1+y^{4}} dt)$
But then I'm not sure if the differential quantity changes, namely:
$\int_{0}^{\sin(x)} 2y\sqrt{1+y^{4}} dt$
or
$\int_{0}^{\sin(x)} 2y\sqrt{1+y^{4}} dy$
I am not allowed to go beyond single variable calculus (so no double integrals allowed) nor am I allowed to use methods taught outside a standard introductory calculus. So I can't use trigonometric substitution, partial fractions, Integration by parts etc...
Any ideas about this?
SECOND EDIT:
So I did the question as follows.
Let $f(x)=\int_{0}^{\sin(x)} \sqrt{1+t^{2}} dt$
Thus we have, $$g(y)=\int_{3}^{y^{2}} f(x)$$
$$g(y)=F(y^2)-F(3)$$
$$g'(y)=2yf(y^2)-0$$
$$g'(y)=2yf(y^2)$$
$$g''(y)=2yf'(y^2)(2y)+2f(y^2)$$
$$g''(y)=4y^2f'(y^2)+2f(y^2)$$
So, $$f(y^2)=\int_{0}^{\sin(y^2)} \sqrt{1+t^{2}} dt$$
and,
$$f'(y^2)=\cos(y^2)\sqrt{1+\sin^{2}(y^2)}$$
Therefore,
$$g''(y)=4y^2f'(y^2)+2f(y^2)$$
$$g''(y)=4y^2\cos(y^2)\sqrt{1+\sin^{2}(y^2)}+2\int_{0}^{\sin(y^2)} \sqrt{1+t^{2}}$$
$$g''(\sqrt{\pi})=-4\pi$$

You can first integrate inner integral which has a known integration. After integrating ,putting limits  it becomes $$\frac {\sin (x)}{2}\sqrt {1+\sin^2 (x)}+\frac {1}{2}\ln (\sin (x)+\sqrt {1+\sin^2 (x)}) $$ and then you have integral wrt x. And now use fundamental theorem of calculus .
