area under a parametric curve problem 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?
 A: 
The parametric curve in question here is one of the simpler Lissajous figures, for which "running" the parameter from $ \ t = 0 \ \ \text{to} \ \ t = \frac{\pi}{2} \ $ traces the portion in the first quadrant.  So finding the area "under the curve" for this range of the parameter calls for the integration to be carried out in the first quadrant.  Since the curve is traced "backwards" (right to left), the usual sense for the area integration will be recovered by introducing a minus sign:
$$ \int_0^1 \ y(x) \ \ dx \ \ \rightarrow \ \  \int_{\pi / 2}^0 \ \sin \ 2t \ \cdot \  (- \sin \ t \ \ dt) \ \ = \ \ - \int^{\pi / 2}_0 \ \sin \ 2t \ \cdot \  (- \sin \ t \ \ dt) $$
$$ = \ \  \int^{\pi / 2}_0 \ \sin \ 2t \  \sin \ t \ \ dt \ \ . $$
Applying the "product-to-sum" formulas, we find this area as
$$ = \ \  \int^{\pi / 2}_0 \ \frac{1}{2} \ [ \ \cos \ t \ - \  \cos \ 3t \ ] \ \ dt \ =  \ \frac{1}{2}  \left[ \ \sin \ t \ - \  \frac{1}{3} \sin \ 3t \ \right] \vert^{\pi / 2}_0  $$
$$ =  \ \frac{1}{2}  \left[ \ ( \sin  \frac{\pi}{2} \ - \  \frac{1}{3} \sin  \frac{3\pi}{2} ) \ - \ ( \sin \ 0 \ - \  \frac{1}{3} \sin  \ 0 ) \ \right]   $$
$$ =  \ \frac{1}{2}   \ ( \ 1 \ - \  \frac{1}{3}  \cdot  [-1] \ )  \ = \ \frac{2}{3} \ \ ,   $$
with the area enclosed by the complete curve being four times this.
