Question

求xsinxcosx的不定积分

Answer 1

原式=(1/2)∫xsin2xdx
令t=2x, dx=(1/2)dt
原式=(1/8)∫tsintdt
=(-1/8)∫td(cost)
=(-1/8)tcost+(1/8)∫costdt
=(-1/8)tcost+(1/8)sint+C
=(-1/4)xcos2x+(1/8)sin2x+C