求不定積分∫(e^x)sin²xdx
解：原式=(1/2)∫(e^x)(1-cos2x)dx
=(1/2)[(e^x)-∫(e^x)cos2xdx]
=(1/2)[e^x-∫cos2xd(e^x)]
=(1/2)e^x-(1/2)
[(e^x)cos2x+2∫(e^x)sin2xdx]
=(1/2)(1-cos2x)(e^x)-∫(e^x)sin2xdx
=(1/2)(1-cos2x)(e^x)-∫sin2xd(e^x)
=(1/2)(1-cos2x)(e^x)-[(sin2x)(e^x)-2∫(e^x)cos2xdx]
=(1/2)(1-cos2x)(e^x)-(sin2x)(e^x)+2∫(e^x)cos2xdx
移項得
(5/2)∫(e^x)cos2xdx
=(1/2)e^x-(1/2)(1-cos2x)(e^x)+(sin2x)(e^x)
=(1/2)(cos2x+2sin2x)(e^x)
故∫(e^x)cos2xdx
=(1/5)(cos2x+2sin2x)(e^x)
故原式
=(1/2)e^x-(1/5)(cos2x+2sin2x)(e^x)+C
=[(1/2)-(1/5)(cos2x+2sin2x)]e^x+C.
求不定積分∫(e^x)sin²xdx
解：原式=(1/2)∫(e^x)(1-cos2x)dx
=(1/2)[(e^x)-∫(e^x)cos2xdx]
=(1/2)[e^x-∫cos2xd(e^x)]
=(1/2)e^x-(1/2)
[(e^x)cos2x+2∫(e^x)sin2xdx]
=(1/2)(1-cos2x)(e^x)-∫(e^x)sin2xdx
=(1/2)(1-cos2x)(e^x)-∫sin2xd(e^x)
=(1/2)(1-cos2x)(e^x)-[(sin2x)(e^x)-2∫(e^x)cos2xdx]
=(1/2)(1-cos2x)(e^x)-(sin2x)(e^x)+2∫(e^x)cos2xdx
移項得
(5/2)∫(e^x)cos2xdx
=(1/2)e^x-(1/2)(1-cos2x)(e^x)+(sin2x)(e^x)
=(1/2)(cos2x+2sin2x)(e^x)
故∫(e^x)cos2xdx
=(1/5)(cos2x+2sin2x)(e^x)
故原式
=(1/2)e^x-(1/5)(cos2x+2sin2x)(e^x)+C
=[(1/2)-(1/5)(cos2x+2sin2x)]e^x+C.