定积分怎么化成的？
展开全部
I = ∫(-π/4->π/4) (sinx)^2/(1+e^x) dx
u=-x
x=-π/4 , u=π/4
x=π/4 , u=-π/4
I
= ∫(-π/4->π/4) (sinx)^2/(1+e^x) dx
= ∫(π/4->-π/4) (sinu)^2/(1+e^(-u)) (-du)
= ∫(-π/4->π/4) (sinx)^2/(1+e^(-x)) dx
= ∫(-π/4->π/4) e^x. (sinx)^2/(1+e^x) dx
2I
= ∫(-π/4->π/4) (sinx)^2/(1+e^x) dx +∫(-π/4->π/4) e^x. (sinx)^2/(1+e^x) dx
= ∫(-π/4->π/4) (sinx)^2 dx
u=-x
x=-π/4 , u=π/4
x=π/4 , u=-π/4
I
= ∫(-π/4->π/4) (sinx)^2/(1+e^x) dx
= ∫(π/4->-π/4) (sinu)^2/(1+e^(-u)) (-du)
= ∫(-π/4->π/4) (sinx)^2/(1+e^(-x)) dx
= ∫(-π/4->π/4) e^x. (sinx)^2/(1+e^x) dx
2I
= ∫(-π/4->π/4) (sinx)^2/(1+e^x) dx +∫(-π/4->π/4) e^x. (sinx)^2/(1+e^x) dx
= ∫(-π/4->π/4) (sinx)^2 dx
本回答由提问者推荐
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
