106something similar but not quite:
in that question, the angle was in the GINT portion , here the sin finction is in the GINT.
as sinx is periodic with period 2\pi, so the integral can be written as
(n-1)\int_{0}^{2\pi }{[sinx]dx} + \int_{\pi }^{2\pi }{[sinx]dx}
Consider the function f(x) = [sinx]
when x = 0 to pi, the function is zero except at x=pi/2 where f(x) = 1. But f(x) being 1 at one point does not contribute to the integral (in case n is finite) So essentially
\int_{0}^{\pi }{[sinx]dx} =0
In this case the integral can be rewritten as
n\int_{\pi }^{2\pi }{[sinx]dx} =0
consider f(x) when x=pi to x=2pi.
Then f(x) takes the value -1 except at x=3pi/2 when x=-2. Again f(x) being -2 at one point does not contribute to the integral (in case n is finite)
So [sinx] in pi to 2pi = -1
So the integral becomes
n\int_{\pi }^{2\pi }{-1dx} =0
= -n\pi