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).` **

## We’ll help your grades soar

Start your 48-hour free trial and unlock all the summaries, Q&A, and analyses you need to get better grades now.

- 30,000+ book summaries
- 20% study tools discount
- Ad-free content
- PDF downloads
- 300,000+ answers
- 5-star customer support

Already a member? Log in here.

Are you a teacher? Sign up now