Integration by parts (2-x)cos(nPi/2)x

Hi,

I'm doing fouier transforms and I'm not sure how to integrate (2-x)cos(nPi/2)x, (-Pi,Pi). Anyone able to help me out? Even the indefinite integral would be fine.

I guess u would be (2-x) and dv would be cos(nPi/2)x dx. I'm not sure how to handle (nPi/2) since it is not a constant. Does (nPi/2) stay with x at all times?

Thanks

npi/s is a constant for the purposes of the integration (in which the only variable is x). So, after making
u = 2-x and dv=cos((npi/2)x) we get
(2/(npi)) (2-x) sin((npi/2)x) + (2/(npi)) Int sin((npi/2)x) dx = (2/(npi)) (2-x) sin((npi/2)x) - (2/(npi))^2 cos((npi/2)x) between -pi and pi, which gives us
(2/(npi)) (2-pi) sin (n pi^2/2) + (2/(npi)) (2+pi) sin (n pi^2/2) - (2/(npi))^2 [ cos (npi^2/2) - cos (-npi^2/2) ] =
(8/(npi)) sin (n pi^2 /2)