= Solution
Let $a=\pi/3$, $b=2\pi/3$ and take $n$ equal subintervals of width $h=(b-a)/n=\pi/(3n)$. Since <sine> is continuous, its integral is the limit of the left-endpoint <Riemann sums>
$$
J=\lim_{n\to\infty}h\sum_{j=0}^{n-1}\sin(a+jh).
$$
Using <Euler's formula>, the corresponding complex sum is a finite <geometric series>:
$$
S_n=h\sum_{j=0}^{n-1}e^{i(a+jh)}=he^{ia}\frac{1-e^{inh}}{1-e^{ih}},\qquad nh=b-a=\frac\pi3.
$$
Since $(e^{ih}-1)/h\to i$,
$$
\lim S_n=\frac{e^{ib}-e^{ia}}i=\frac{(-1/2+i\sqrt3/2)-(1/2+i\sqrt3/2)}i=i.
$$
Taking <imaginary parts> gives
$$
\boxed{J=1.}
$$
One can see the same limit entirely within the real sums: the finite trigonometric-sum identity obtained from this geometric sum gives
$$
h\sum_{j=0}^{n-1}\sin(a+jh)=h\frac{\sin(nh/2)}{\sin(h/2)}\sin\left(a+\frac{n-1}{2}h\right)=\frac h2\cot\frac h2\longrightarrow1.
$$
Thus the evaluation is explicitly by a <Riemann sum>, as required.
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