# Thread: integration problem - antiderivative??

1. ## integration problem - antiderivative??

how is the integral of sin 2t = -1/2cos 2t?

i cant seem to get where the 1/2 comes from.

2. $\displaystyle \int{\sin{a}x} = -\dfrac{1}{a}\cos{a}x$.

3. It might help to look at the problem in reverse:

$\displaystyle \frac{d}{dt}\left(-\frac{1}{2}\cos{2t}\right)=-\frac{1}{2}(-sin{2t})\frac{d}{dt}(2t)=sin{2t}$

The 1/2 is there to cancel the 2 coming from the $\displaystyle \frac{d}{dt}(2t)$ part of the chain rule.

Post again in this thread if you're still having trouble.

4. Originally Posted by calculus0
how is the integral of sin 2t = -1/2cos 2t?

i cant seem to get where the 1/2 comes from.
If you do not understand the direct formula, try substitution method:

let $\displaystyle u = 2t$ then $\displaystyle du = 2 dt \rightarrow dt = \frac{du}{2}$

$\displaystyle \int sin2t dt = \int sin(u)\frac{du}{2} = \frac{1}{2} (-cos(u))+C$

substitute $\displaystyle u = 2t$ in the above answer"

$\displaystyle - \frac{1}{2} cos2t+C$