$F(x)=\int_{\pi}^{x} \frac{e^t}{3+\sin t} dt$ Find $(F^{-1})'(0)$

Question

Let $F$ be a function: $F(x)=\int_{\pi}^{x} \frac{e^t}{3+\sin t} dt$

Find $(F^{-1})'(0)$.

What I've been doing:

What I tried to do was to find $F'(x)$ and $F^{-1}(0)$ such that I could use $\frac{1}{F'(F^{-1}(x))}$

I found that $F'(x)=\frac{e^x}{3+\sin x}$, however when I try to integrate the original function I can't. I tried doing it by substitution, with $u=e^x$ or $u=3+\sin x$.

Is there something I'm missing? Do I need to integrate when trying to find $(F^{-1})'(0)$?

Answer 1

All you need to do is\begin{align}(F^{-1})'(0)&=\frac1{F'\bigl(F^{-1}(0)\bigr)}\\&=\frac1{F'(\pi)}\\&=\frac{3^\pi}{1+\sin\pi}\\&=3^\pi.\end{align}