# int_0^(1/2) x cos(pi x) dx Evaluate the integral

## Expert Answers 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).

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