(x^2-1)cos2xdx
=∫(x²-1)cos2xdx
=(1/2)∫(x²-1)d(sin2x)
=(1/2)[(x²-1)sin2x-∫2xsin2xdx]
=(1/2)[(x²-1)sin2x+∫xd(cos2x)]
=(1/2)[(x²-1)sin2x+xcos2x-∫cos2xdx]
=(1/2)[(x²-1)sin2x+xcos2x-(1/2)sin2x]+C
=(1/2)[(x²-3/2)sin2x+xcos2x]+C
(x^2-1)cos2xdx,求不定积分
(x^2-1)cos2xdx
=∫(x²-1)cos2xdx
=(1/2)∫(x²-1)d(sin2x)
=(1/2)[(x²-1)sin2x-∫2xsin2xdx]
=(1/2)[(x²-1)sin2x+∫xd(cos2x)]
=(1/2)[(x²-1)sin2x+xcos2x-∫cos2xdx]
=(1/2)[(x²-1)sin2x+xcos2x-(1/2)sin2x]+C
=(1/2)[(x²-3/2)sin2x+xcos2x]+C