(1-t)(cosnπt) 對t積分,積分區間0->1

_∫(1-t)cos(nπt)dt=∫[cos(nπt)-tcos(nπt)]dt=∫cos(nπt)dt-∫tcos(nπt)dt=sin(nπt)/(nπ)-∫tcos(nπt)dt ---(1) _∫tcos(nπt)dtPut u=t,du=dt.Put dv=cos(nπt)dt, v=sin(nπt)/(nπ).Sub. this into ∫udv=uv-∫vdu.=tsin(nπt)/(nπ)-∫sin(nπt)/(nπ)dt=tsin(nπt)/(nπ)+cos(nπt)/(nπ)² ---(2) Sub. (2) into (1)._∫(1-t)cos(nπt)dt=sin(nπt)/(nπ)-[tsin(nπt)/(nπ)+cos(nπt)/(nπ)²]=sin(nπt)/(nπ)-tsin(nπt)/(nπ)-cos(nπt)/(nπ)²=[nπ(1-t)sin(nπt)-cos(nπt)]/(nπ)² So,_∫[0~1](1-t)cos(nπt)dt=[[nπ(1-t)sin(nπt)-cos(nπt)]/(nπ)²][0~1]=[nπ(0)sin(nπ)-cos(nπ)]/(nπ)²-(nπsin0-cos0)/(nπ)²=-cos(nπ)/(nπ)²+1/(nπ)²=[1-cos(nπ)]/(nπ)²