# area under a parametric curve problem

so 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 therefore cancels the sign we get by differentiating cos(t), and he winds up with a positive integral like this:

$$\int_0^{pi/2} sin(2t)sin(t)dt$$

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?

-
I don't understand. You have a minus sign too. how does your answer differ from the teacher's? And what's "cost"? By the way, using "\sin" gives you $\sin$ and is a lot easier on the eyes. –  Stefan Smith Nov 14 '13 at 3:17
He starts the ingtegral with a sign, and the differentiation cancels the two signs.. I am left with a sign. I don't really see what's unclear about basic cancellation. –  Scuba Steve Nov 14 '13 at 3:19