What is the velocity of a particle at t=2seconds given force function (x)

[aitee]

Homework Statement

Force Fx = (10N) sin( (2*pi*t)/4 ) where t is in seconds

this force Fx is exerted on a 210 g particle during the interval 0s <= t <= 2s


If the particle starts from rest, what is its speed at t = 2.0s ?

Homework Equations

F=ma

I think this is the only relevant equation.

The Attempt at a Solution

So I know that Force = Mass * Accel, therefore Accel = F/m. So because I am given the force as a function and have the mass I basically have the acceleration function.. Is this correct so far? Because velocity is the integral of acceleration, I can get velocity with that.
I graphed it online (for visual purposes), calculated the integral from 0 to 2 seconds and got the answer.

I'm having trouble doing it on paper. Simplifying a = F/m, I get: a = 10(sin (pi/2)*t )/ .21
Integrating this I pull out the 10/.21 constant,
the integral of sin( (t*pi) /2 ) is: -cos( (t*pi)/2 ) * (pi/2) evaluated at 2 and 0. (that second pi/2 is correct by chain rule?)

I'm evaluating this as -cos( (2pi)/2 )* (pi/2) - -cos(0) (pi/2)

Both cos(0) and cos(pi) == 1

so: -(pi/2) + (pi/2) = 0

Multiplied by that constant part on the outside (10/.21) is still zero..

I don't see why this is wrong.. The link below at wolframalpha.com shows what I used to get the answer:
http://www.wolframalpha.com/input/?i=integral+of+(10*sin((pi*x)/2)/.21)+from+0+to+2


Any help on how to solve this on paper (the steps) or help finding my stupid mistake is would be appreciated.

 

[rl.bhat]

the integral of sin( (t*pi) /2 ) is: -cos( (t*pi)/2 ) * (pi/2)

It should be -cos(πt/2)*2/π