Determine the extrema of $\int^{x^2}_1\frac{\sin t}{2+e^t}\,dt$

Determine the local extrema of $F(x)=\int^{x^2}_1\frac{\sin t}{2+e^t}\,dt$.
Prove that $|F(x)|\leq|x-1|$ for all $x$.

$$F'(x)=\frac{2x\sin x}{2+e^x}\mbox{,}$$ so that the set of all points at which $F$ may attains its local extremum are $n\pi$, $n\in\mathbb{Z}$.

I don't know how to proceed any further.

The maybe more than one way to answer your question, however using the mean value theorem would be the most direct way to do it.

We have, $$ |F'(x)|=\left|\dfrac{2x\sin(x^2)}{2+\exp(x^2)}\right|\leq\left|\dfrac{2x}{2+\exp(x^2)}\right|\leq\left| \dfrac{x}{1 + \frac{1}{2}\exp(x^2)}\right|\leq 1$$ The last inequality is valid for all $x\in\mathbb{R}$, because then we have $2x\leq\exp(x^2)$

Given that $F(x)$ is continuous on $[1,x]$ and derivable on $]1,x[$, we can conclude using the MVT: $$ \left|\dfrac{F(x) - F(1)}{x-1} \right|\leq 1$$ Finall, given that $F(1)=0$ we have $$ \left| F(x)\right| \leq \left| x-1 \right|, \forall x\in\mathbb{R}$$

About the local extrema: To find the local extrema of the function $F$, we need to solve the equation $F'(x)=0 $.

$$ F'(x)=0 \Longleftrightarrow \frac{2x\sin x^2}{2+\exp(x^2)}=0$$ Thus, $$ x= 0,\quad \mbox{or }\ \sin(x^2)=0 \Leftrightarrow x^2=n\pi,\ n\in\mathbb{Z}$$

We can than conclude that the local extrema of the function are of the form $\pm \sqrt{n\pi}, n \in \mathbb{N}$