So one may prove the second equality of the question (the so-called Weyl integral formula) in the following way:

So we assume that $G$ is endowed with a bi-invariant metric with total volume $1$. For an element $g\in G$ with denote by $G(g)$ the conjugacy class of $g$ in $G$. We say that $g$ is generic if $dim(G(g))=dim(G)-rank(G)=m$. We choose a maximal torus $T$ and we denote by $W$ the Weyl group which acts on it. Recall that every generic conjugacy class of $G$ intersects orthogonally $T$ at $|W|$-points. Moreover, $T$ is a totally geodesic submanifold of $G$. For every $H\in\mathfrak{h}$ we denote $$ Q(H):=\prod_{\alpha\in\Delta^+}(e^{\pi i(\alpha,H)}-e^{-\pi i(\alpha,H))}). $$ We think of the roots of $G$ as elements of a fixed Weyl chamber (which we denote by $C_0$) of $\mathfrak{h}:=Lie(T)$. Recall finally that $\mathfrak{h}$ is endowed with an $ad$-invariant inner product for which $W$ acts by isomoetries. Finally we denote by $v(t)$ the "$m$ dimensional volume" of the conjugacy class $G(t)$ for a generic element $t\in T$.

(1) First for a generic element $t\in T$ such that $exp(H)=t$ one may show that that $$ v(t)=c\prod_{\alpha\in\Delta^+} 4sin^2(\pi(\alpha,H)) $$ for some real number $c$ which has to be determined.

(2) The Weyl group $W$ is generated by reflexions $r_j$. For every $r_j$ there is an associated simple root $\alpha_j$ such that $r_j(\Delta^+)=(\Delta^+-\{\alpha_j\})\cup\{-\alpha_j\}$. Using this observation one sees that $Q(r_jH)=-Q(H)$ and hence $Q(\sigma H)=sign(\sigma)Q(H)$. Now since $Q(H)$ is an alternating function with respect to the $W$-action one has that $$ Q(H)=\sum_{\sigma\in W}sign(\sigma)e^{2\pi i(\sigma(\delta),H)}+\mbox{possible other terms}. $$ Note that $\delta$ is the only vector of the form $\frac{1}{2}\sum \pm\alpha$ which is in $C_0$. It follows from this that there is in fact no other terms!

(3) Finally one shows that $c=1$ by observing that the set of functions $\{H\mapsto e^{2\pi i(\sigma(\delta),H)}:\sigma\in W\}$ are orthonormal in $L^2(T)$.

So one may prove the second equality of the question (the so-called Weyl integral formula) in the following way:

For every $H\in\mathfrak{h}=Lie(T)$ we denote $$ Q(H):=\prod_{\alpha\in\Delta^+}(e^{\pi i(\alpha,H)}-e^{-\pi i(\alpha,H))}). $$ We think of the roots of $G$ as elements of a fixed Weyl chamber (which we denote by $C_0$) of $\mathfrak{h}$.

The Weyl group $W$ is generated by reflexions $r_j$. For every $r_j$ there is an associated simple root $\alpha_j$ such that $r_j(\Delta^+)=(\Delta^+-\{\alpha_j\})\cup\{-\alpha_j\}$. Using this observation one sees that $Q(r_jH)=-Q(H)$ and hence $Q(\sigma H)=sign(\sigma)Q(H)$. Now since $Q(H)$ is an alternating function with respect to the $W$-action one has that $$ Q(H)=\sum_{\sigma\in W}sign(\sigma)e^{2\pi i(\sigma(\delta),H)}+\mbox{possible other terms}. $$ Now the key observation is to note that $\delta$ is the only vector of the form $\frac{1}{2}\sum \pm\alpha$ which is in $C_0$. It follows from this that there is in fact no other terms!