Calculus: Early Transcendentals, Chapter 7, 7.1, Section 7.1, Problem 23

You need to solve the integral int_0^(1/2) (x) cos (pi*x) dx , hence, you need to use substitution pi*x = t => pi*dx = dt => dx = (dt)/(pi)
int x*cos (pi*x) dx = 1/(pi^2) int t*cos t
You need to use the integration by parts for int t*cos t such that:
int udv = uv - int vdu
u = t => du = dt
dv =cos t=>v =sin t
int t*cos t = t*sin t- int sin t dt
1/(pi^2) int t*cos t = 1/(pi^2)(t*sin t +cos t) + c
Replacing back the variable yields:
int x*cos (pi*x) dx = 1/(pi^2)(pi*x*sin(pi*x) +cos (pi*x)) + c
Using the fundamental theorem of calculus, yields:
int_0^(1/2) (x)cos (pi*x) dx = 1/(pi^2)(pi*(1/2)*sin(pi/2) +cos (pi/2) - 0*sin 0 - cos 0)
int_0^(1/2) (x)cos (pi*x) dx = 1/(pi^2)(pi/2 - 1)
Hence, evaluating the integral, using integration by parts, yields int_0^(1/2) (x)cos (pi*x) dx = 1/(pi^2)(pi/2 - 1).