Integral of $-4\sin(2t - (pi/2)) $ weird behavior on wolfram alpha I'm confused by what Wolfram Alpha is doing with my function:
$$-4\sin{(2t - (\pi/2))}$$
on why the it gets replaced by 
$$4\cos{(2t)}$$. 
Is it equal? Link: See behavior here

Yes. Using the angle-addition formula,
$$ \sin{(2t-\pi/2)} = \sin{2t}\cos{(\pi/2)}-\cos{2t}\sin{(\pi/2)} = -\cos{2t}. $$
