求一道定积分 ∫(π,0)xsin2x dx

2024-11-16 09:12:44
推荐回答(1个)
回答1:

∫xsin2xdx
=1/2*∫xsin2xd2x
=-1/2*∫xdcos2x
=-1/2*xcos2x+1/2*∫cos2xdx
=-1/2*xcos2x+1/4*∫cos2xd2x
=-1/2*xcos2x+1/4*sin2x+C
把积分限代入
所以原式=-π/2