I have a parametric curve, $$x = \cos(t),$$ $$y = \sin(2t).$$
I found that I need the area from 0 to $\pi/2.$
put this into an integral in terms of $t$, I get
$$ -\int_0^{\pi/2} \sin(2t) \sin(t)dt $$
But in my teacher's solution, he starts the integral off with a negative, and cancels the sign we get by differentiating cos(t). Therefore he winds up with a positive integral like the following:
$$ \int_0^{\pi/2} \sin(2t)\sin(t)dt. $$
I really cannot see how he's justified the extra negative sign.
Can anyone tell me how he's gone about solving it this way?