# How do you evaluate the definite integral #int(-sinx)dx# from #[pi/6,pi/2]#?

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To evaluate the definite integral (\int_{\frac{\pi}{6}}^{\frac{\pi}{2}} -\sin(x) , dx), you need to integrate the function (-\sin(x)) with respect to (x) over the interval ([\frac{\pi}{6}, \frac{\pi}{2}]).

The integral of (-\sin(x)) is (\cos(x)), so integrating from (\frac{\pi}{6}) to (\frac{\pi}{2}) gives:

[ \cos\left(\frac{\pi}{2}\right) - \cos\left(\frac{\pi}{6}\right) = 0 - \frac{\sqrt{3}}{2} = -\frac{\sqrt{3}}{2} ]

Therefore, the value of the definite integral is (-\frac{\sqrt{3}}{2}).

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