I have been working on this question for a while and my answer is not correct. The question is to evaluate this integral in terms of sums without absolute values. Here's what I have, but the answer is not correct and I can't seem to catch why:
$\int_{\pi/3}^{\pi} \vert \cos x \vert dx$
$ = \int_{\pi/3}^{\pi/2} \cos x + \int_{\pi/2}^{\pi} -\cos x$
$ = [\cos(\pi/2) - \cos(\pi/3)] + [-\cos(\pi) - (-\cos(\pi/2))]$
$ = -\cos(\pi/3) + 1$
$ = 0.5$
**Sorry for the formatting but when I type 'pi' it doesn't change to the symbol
The primitive of $\cos$ is $\sin,$ not $\cos$ again.
That is, you should have $$\int{ \cos x \mathrm d x}=\sin x + c.$$