sin^2cos^4求定积分,区间是【0,π/2】,^是几次方.
展开全部
∫(sinx)^2(cosx)^4dx
=1/4∫(2sinxcosx)^2(cosx)^2dx
=1/4∫(sin2x)^2(cosx)^2dx
=1/8∫(sin2x)^2(1+cos2x)dx
=1/8∫(sin2x)^2dx+1/8∫(sin2x)^2cos2xdx
=1/16∫(1-cos4x)dx+1/16∫(sin2x)^2d(sin2x)
=x/16-sin4x/64+(sin2x)^3/48+C
∫(下限为0 上限为π/2)(sinx)^2(cosx)^4dx
=π/32-sin2π/64+(sinπ)^3/48-[0-sin0/64+(sin0)^3/48]
=π/32
=1/4∫(2sinxcosx)^2(cosx)^2dx
=1/4∫(sin2x)^2(cosx)^2dx
=1/8∫(sin2x)^2(1+cos2x)dx
=1/8∫(sin2x)^2dx+1/8∫(sin2x)^2cos2xdx
=1/16∫(1-cos4x)dx+1/16∫(sin2x)^2d(sin2x)
=x/16-sin4x/64+(sin2x)^3/48+C
∫(下限为0 上限为π/2)(sinx)^2(cosx)^4dx
=π/32-sin2π/64+(sinπ)^3/48-[0-sin0/64+(sin0)^3/48]
=π/32
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
