Integrate [0, 1/2] xcos (pi*x).

To answer your question, this could be the possible answer and i hope you understand and interpret it correctly:

[Integrate [0, 1/2] xcos (pi*x

let u=x so that du=dx

and v=intgral cos (xpi) dx

v = (1/pi) sin (pi*x)

integration by parts

uv-itgral[0,1/2]vdu just plug ins

(1/pi) sinpi*x] - (1/pi) itgrlsin (pi*x) dx from 0 to 1/2

(1/pi) x sinpi*x - (1/pi) [ - (1/pi) cos pi*x] from 0 to 1/2

= (1/2pi) + (1/pi^2) [cos pi*x/2-cos 0]

=1/2pi - 1/2pi^2

= (pi-2) / 2pi^2 ans