I used $u$-substitution to break down the problem to be $-\cos(u)$ and then got my final answer to be $-\cos(x-\pi/3)$ (re-insertion for my $u$) and it is not completely correct. What am I missing or is there another step?
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$\begingroup$Your solution is completely right up to integration constant.
$$I=\int \sin\left(x-\frac{\pi}{3}\right)\,dx$$
Substitution $u=x-\frac{\pi}{3}$, $du=dx$:
$$I=\int \sin\left(u\right)\,du=-\cos\left(u\right)+C=-\cos\left(x-\frac{\pi}{3}\right)+C$$
$\endgroup$ $\begingroup$Since $\cos(u) =\sin(\pi/2-u) $ and $\cos(u) =\cos(-u) $, $\cos(x-pi/3) =\cos(\pi/3-x) =\sin(\pi/2-(\pi/3-x)) =\sin(\pi/6+x) $.
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