Calculus help integration by parts

How do I solve this?
integral of xcos(πx)dx from 0 to 1/2
answer is (π-2)/(2π^2)
I tried to do integration by parts and ended up with (1/π)sin(πx)x+(1/π^2)(cos(πx) from 0 to 1/2

When you evaluate your answer at 1/2, you get:

1/pi * 1 * 1/2 + 0

When you evaluate at 0, you get

1/pi^2

So 1/2pi - 1/pi^2 = pi/2pi^2 - 2/2pi^2 = (pi-2)/2pi^2

So you integrated correctly! you just need to substitute in.

1/2
∫ xcos(πx)dx
0

let u = x....dv = cos(πx)
du = dx.....v = 1/πsin(πx)

∫ udv = uv - ∫ vdu

= 1/πxsin(πx) - 1/π ∫sin(πx)dx

.......................................… 1/2
= 1/πxsin(πx) + 1/π^2cos(πx)|
.......................................… 0


={1/π(1/2)sin[π(1/2)] + 1/π^2cos[π(1/2)]} - {1/π(0)sin[π(0)] + 1/π^2cos[π(0)]}
=[1(2π) + 1/π^2(0)] - 1
= [1/(2π) - 1/π^2]2π^2
...π - 2
=------------- answer//
.....2π^2

Evaluate your answer at the limits of integration.

(1/π)sin(π*0)*0+(1/π^2)(cos(π*0) =1/π^2
(1/π)sin(π/2)/2+(1/π^2)(cos(π/2) = 1/(2π)

1/(2π) - 1/π^2 = (π - 2)/(2π^2)

∫xcos(πx) dx from 0 to 1/2

u = x => du = dx

dv = cos(πx) dx => v = 1/(π) * sin(πx)

uv - ∫v du

x*sin(πx)/π + 1/(π²) * cos(πx) eval. from 0 to 1/2

= 1/(2π) * sin(π/2) + 1/(π²) * cos(π/2) - 1/π²

= 1/(2π) - 1/π² = (π - 2)/(2π²)

Your integral is correct.
Now evaluate from 0 to 1/2
[sin(pi/2)/2pi+0] - [0+cos (0)/pi^2] =
1/2pi - 1/pi^2] =
pi/2pi^2 - 2/2pi^2 =
(pi-2)/2pi^2

You did the integration correctly; just evaluate it at x = 1/2 and subtract from the the your results evaluated at x =0.

u = x du = dx dv = cos[pix]dx v = sin[pix]/pi

I = xsin[pix]/pi - I sin[pix]/pi]

I = xsin[pix]/pi +cos[pix]/pi^2 0k u r correct!

That's right. Can't you substitute x = 1/2 and x = 0 without messing up the algebra?