求不定积分 {[(x^2)*arcsinx+1]/√[1-(x^2)]}dx
展开全部
原式=∫(x^2*arcsinx+1)d(arcsinx)=∫x^2*arcsinxd(arcsinx)+∫d(arcsinx)=∫x^2*arcsinxd(arcsinx)+arcsinx
令x=sint(t属于[-π/2,π/2])
则原式=∫tsin^2(t)dt+t=∫t(1-cos(2t))/2dt+t=∫t/2dt-1/2∫tcos(2t)dt+t=1/4t^2-1/4∫td(sin(2t))+t=1/4t^2-1/4tsin(2t)+1/4∫sin(2t)dt+t=1/4t^2-1/4tsin(2t)-1/8cos(2t)+t=1/4t^2-1/2tsintcost-1/8+1/4sin^2(t)+t=1/4arcsin^2(x)-1/2x*√(1-x^2)*arcsinx-1/8+1/4x^2+arcsinx
令x=sint(t属于[-π/2,π/2])
则原式=∫tsin^2(t)dt+t=∫t(1-cos(2t))/2dt+t=∫t/2dt-1/2∫tcos(2t)dt+t=1/4t^2-1/4∫td(sin(2t))+t=1/4t^2-1/4tsin(2t)+1/4∫sin(2t)dt+t=1/4t^2-1/4tsin(2t)-1/8cos(2t)+t=1/4t^2-1/2tsintcost-1/8+1/4sin^2(t)+t=1/4arcsin^2(x)-1/2x*√(1-x^2)*arcsinx-1/8+1/4x^2+arcsinx
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