Forgot how to integrate yes t*cos(Pi*t)

[mr_coffee]

forgot how to integrate! yes! t*cos(Pi*t)

Hello everyone I'm integrating a position vector and I'm stuck on integrating the j unit. t*cos(Pi*t);
the answer i got with maple is:
1/Pi^2*(cos(Pi*t)+Pi*t*sin(Pi*t))
but i have no idea how maple busted that out.
if i let u = cos(Pi*t);
du = sin(Pi*t)*Pi dt;
1/Pi du = sin(Pi*t);
but i don't see how this is helping me any...

 

[TD]

Use integration by parts: f = t and dg = cos(pi*t)dt

Then $\int {fdg = fg - } \int {gdf}$

 

[mr_coffee]

Thanks for the responce but I'm still messing it up!
I let f = t; dg = cos(Pi*t) dt;
df = 1;
i integrated dg, to get g, and got:
g = [t*sin(Pi*t)]/Pi;

then u said:
fg - integral(g*df);
(1)([t*sin(Pi*t)]/Pi) - integral (t*sin(Pi*t)]/Pi)(1); but now I'm stuck integrating this function by parts too?

 

[sqrt(-1)]

Your integral for dg has found a factor of t for some reason, your integral should be:

$$\int \cos (\pi t ) dt = \frac{1}{\pi} \sin(\pi t)$$

 

[mr_coffee]

i had that, but the def says: $\int {fdg = fg - } \int {gdf}$ so doesn't this mean i have to take f which is t, and multiply it by g? which is $$\int \cos (\pi t ) dt = \frac{1}{\pi} \sin(\pi t)$$ that's where i got that t from

 

[sqrt(-1)]

Remember the integral on the RHS is asking for the derivative of f, so we have

$$\int t\cos(\pi t) dt = \frac{t}{\pi}\sin(\pi t) - \frac{1}{\pi} \int \sin( \pi t ) dt$$

 

[mr_coffee]

ohhh thanks again sqrt!