高数题定积分计算
展开全部
令t=arccosx，则x=cost，dx=-sintdt
原式=∫(π/2,0) cos^2t*t/sint*(-sint)dt
=∫(0,π/2) tcos^2tdt
=(1/2)*∫(0,π/2) t(1+cos2t)dt
=(1/2)*∫(0,π/2) tdt+(1/2)*∫(0,π/2) tcos2tdt
=(1/4)*t^2|(0,π/2)+(1/4)*∫(0,π/2) td(sin2t)
=π^2/16+(1/4)*tsin2t|(0,π/2)-(1/4)*∫(0,π/2) sin2tdt
=π^2/16+(1/8)*cos2t|(0,π/2)
=(π^2)/16-1/4
原式=∫(π/2,0) cos^2t*t/sint*(-sint)dt
=∫(0,π/2) tcos^2tdt
=(1/2)*∫(0,π/2) t(1+cos2t)dt
=(1/2)*∫(0,π/2) tdt+(1/2)*∫(0,π/2) tcos2tdt
=(1/4)*t^2|(0,π/2)+(1/4)*∫(0,π/2) td(sin2t)
=π^2/16+(1/4)*tsin2t|(0,π/2)-(1/4)*∫(0,π/2) sin2tdt
=π^2/16+(1/8)*cos2t|(0,π/2)
=(π^2)/16-1/4
本回答被提问者和网友采纳
已赞过
已踩过<
评论
收起
你对这个回答的评价是？
推荐律师服务：
若未解决您的问题，请您详细描述您的问题，通过百度律临进行免费专业咨询
