Let $X=\mathcal{C}([0,1])$ be the Banach space of continuous functions on $[0,1]$ (with the supremum norm) and define a map $F:X\rightarrow X$ by $$ F(f)(x)=\int^{x}_{0}\cos\big(f^2(t)\big)dt,\text{ } x\in [0,1]. $$ Show that $F$ is Fréchet differentiable and compute the Fréchet derivative $DF_{|_f}$ for each $f\in X$.
To be Fréchet differentiable means the following: Let $X,Y$ be normed vector spaces, U open in X, and $F:U\rightarrow Y$. Let $x,h\in U$ and let $T:X\rightarrow Y$ be a linear map. Then the limit $$ \underset{h\rightarrow 0}{\lim}\frac{||F(x+h)-F(x)-Th||_Y}{||h||_X}=0$$ exists. We denote $T$ as $DF_{|_x}$.
Making use of $cos(A+\epsilon B)-cosA=-\epsilon BsinA + \mathcal{O}(\epsilon^2)$. we get that \begin{equation} cos\big[(f(x)+h(x))^2\big]=cos(f^2(x))-h(x)[2f(x)+h(x)]sin(f^2(x)) + \mathcal{O}(h^2(x)). \end{equation} Therefore, we have that \begin{align*}F(f+h)(x)-F(f)(x) &=\int^{x}_{0}-h(x)[2f(x)+h(x)]sin(f^2(x))+\mathcal{O}(h^2(x))dx \\ &=\int^{x}_{0}-2f(x)h(x)sin(f^2(x))-h^2(x)sin(f^2(x))+\mathcal{O}(h^2(x))dx \\ &=\int^{x}_{0}-2f(x)h(x)sin(f^2(x))+\mathcal{O}(h^2(x))dx \\ \end{align*} The last line follows since $-h^2(x)sin(f^2(x))$ is a Big-$\mathcal{O}$ of $h^2(x)$.
Now put $DF_{|_f}(h)=\int^{x}_{0}-2f(x)h(x)sin(f^2(x))dx$ as this expression is linear in $h(x).$
Hence, we have that $$ F(f+h)(x)-F(f)(x)-DF_{|_f}(h)=\int^{x}_{0}\mathcal{O}(h^2(x))dx $$
And here I am now stuck. I don't know what to do with the Big-$\mathcal{O}$ notation in the integral. Am I doing this correctly?