计算定积分∫(-1,1)x^2(sinx/(1+x^4)+√(1-x^2))dx
2个回答
展开全部
原式=∫(-1→1)x^2sinx/(1+x^4)+∫(-1→1)x^2√(1-x^2)dx
因为x^2sinx/(1+x^4)是奇函数，所以前面一项为0
令x=sint
则原式=∫(-π/2→π/2)sin^2(t)cost*cos(t)dt
=∫(-π/2→π/2)(1-cos(2t))/2*(1+cos(2t))/2dt
=1/4∫(-π/2→π/2)(1-cos^2(2t))dt
=1/4∫(-π/2→π/2)dt-1/4∫(-π/2→π/2)(1+cos(4t))/2dt
=t/4|(-π/2→π/2)-t/8|(-π/2→π/2)-sin(4t)/32|(-π/2→π/2)
=π/8
因为x^2sinx/(1+x^4)是奇函数，所以前面一项为0
令x=sint
则原式=∫(-π/2→π/2)sin^2(t)cost*cos(t)dt
=∫(-π/2→π/2)(1-cos(2t))/2*(1+cos(2t))/2dt
=1/4∫(-π/2→π/2)(1-cos^2(2t))dt
=1/4∫(-π/2→π/2)dt-1/4∫(-π/2→π/2)(1+cos(4t))/2dt
=t/4|(-π/2→π/2)-t/8|(-π/2→π/2)-sin(4t)/32|(-π/2→π/2)
=π/8
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
