一道定积分题,急!!!

根号下(1-x^2)^3,上限为1,下限为0
2025-06-27 15:16:10
推荐回答(3个)
回答1:

x=sint,t从0到pi/2,原积分为
积分cos^3tcostdt=积分(1+cos2t)^2/4=积分3/8+cos2t/2+cos4t/8=3pi/16

回答2:

利用分部积分法∫ x sin2x dx =-1/2∫ x d(cos2x) =-1/2[xcos2x-∫cos2x dx ] =-1/2xcos2x+1/2∫cos2x dx =-1/2xcos2x+1/4∫cos2x d(2x

回答3:

16/35