Wikipedia has Lagrange multipliers on Banach spaces. Recall that $\mathbb R$ is a Banach space; the original question has it as the codomain of the constraint function. In this case the Lagrange multiplier is a simple multiplier, that is, a linear function $\mathbb R \rightarrow \mathbb R$. I'll write $\lambda(x) : x \mapsto \lambda x$. Also important to keep in mind is that the existence of Langrage multipliers is a necessary condition but not a sufficient one. Therefore calculating a Lagrange multiplier only yields candidate extrema.
In the present example, to satisfy the conditions for the existence of a Langrange multiplier, we need $G(x) = \int f(x) dx - \pi/2$ so that the constraint is of the form $G(x) = 0$. This won't matter in the present case, but it's required in more complicated ones. Thus we have
$$
DH_f(g) = \int_{-1}^1 \frac{ f'(x) }{ \sqrt{1+ f'(x)^2}} g' dx
\qquad
DG_f(g) = \int_{-1}^1 g(x) dx
$$
A local extremal of $H$ at $f_0$ subject to $G(x)=0$ has a Lagrange multiplier $\lambda$ such that
$$
DH_{f_0} = \lambda DG_{f_0}
$$
Equality is that for functions, so these must be true for all arguments $g$. Note that constant functions $f$ have $DH_f=0$ (since the integrand is zero). Constant functions are candidate local extrema for $H$ with respect to any constraint by choosing multiplier $\lambda=0$. In the present case, however, the constraint set contains no constant functions.
If we didn't know the answer, we could integrate the expression for $DH_f$ by parts and thereby calculate a variational derivative for $H$. Since we have a candidate answer, we can skip that, perform the substitution first, and then integrate by parts. I omit the algebra:
$$
f_0(x) = \sqrt{1+x^2}
\qquad
f_0'(x) = \frac{ -x }{ \sqrt{1+x^2} }
\\
DH_{f_0} = \int_{-1}^1 (-x) g' dx = \int_{-1}^1 g(x) \, dx
$$
Choosing $\lambda=1$, we see that $f_0$ is a candidate extremal of $H$ subject to $G=0$.
I'm done, but a full proof that $f_0$ is a minimum is not. Existence of the Lagrange multiplier, as I said above, is not sufficient. One approach to proving minimality would be as follows:
- Compute the variational derivative $\delta H$, that is, $DH_f(g) = \int \delta H_f(x) \, g \, dx$. The multiplier equation is then $\int (\delta H_f(x) - \lambda \circ \delta G_f(x)) g(x) \, dx$, and we have $\delta G = 1$.
- Show that the operator $g \mapsto \int (\delta H_f(x) - \lambda) g(x) \, dx$ is non-degenerate. Anything in a non-trivial eigenspace of $\delta H$ is a candidate extremal, and this step eliminates them.
- Show that $f_0$ is the unique solution within the constraint set to the differential equation $\delta H_f(x) - \lambda = 0$ for any $\lambda$.
