Taking the above plot I'm looking for some intuition how to think about the integral of $sin(x)$, which is $-cos(x)$ (plus some constant that's assumed to be zero for the sake of readbility). The derivative of $sin(x)$ is easily interpreted visually, the slope at $\pi/2$ at the top of the $sin$ curve is zero so $cos(\pi/2)$ shows just that.
What gets my brain in a knot is why $-cos(\pi/2)$ is also zero. Maybe the hole I'm in starts with thinking about the integral as the area under the $sin$ function. This is clearly not zero, neither from 0 to $\pi/2$ nor 'at' $\pi/2$ for any non-zero slivers of x.
I'm probably not thinking about this the right way so the question boils down to what $-cos(x)$ is telling us about $sin(x)$, visually I can't make any sense of it.
I do appreciate the irony that the derivative of $-cos(x)$ is $sin(x)$!
Best Answer
Since $\int_0^x\sin t dt=1-\cos x$, all that $\cos(\pi/2) =0$ means in area terms is that $\int_0^{\pi/2}\sin t dt=1$, or equivalently $\int_{\pi/2}^x\sin t dt=-\cos x$.