∫(-1,1)[e^(-x^2)[in(x+1)/(x-1)]+cosx(sinx)^2]dx=
来源:学生作业帮 编辑:搜狗做题网作业帮 分类:数学作业 时间:2024/07/08 14:22:01
∫(-1,1)[e^(-x^2)[in(x+1)/(x-1)]+cosx(sinx)^2]dx=
![∫(-1,1)[e^(-x^2)[in(x+1)/(x-1)]+cosx(sinx)^2]dx=](/uploads/image/z/8455454-62-4.jpg?t=%E2%88%AB%28-1%2C1%29%5Be%5E%28-x%5E2%29%5Bin%28x%2B1%29%2F%28x-1%29%5D%2Bcosx%28sinx%29%5E2%5Ddx%3D)
∫(-1,1){e^(-x²)[in(x+1)/(1-x)]+cosxsin²x}dx
设f(x)=e^(-x²)[in(x+1)/(1-x)],由于f(-x)=e^(-x²)[ln(-x+1)/(1+x)]=e^(-x²)[ln(x+1)/(1-x)]ֿ¹
=-e^(x²)[ln(x+1)/(1-x)]=-f(x),且x∈[-1,1],故f(x)是奇函数,∴[-1,1]∫e^(x²)[ln(x+1)/(1-x)]dx=0;
∴[-1,1]∫{e^(-x²)[in(x+1)/(1-x)]+cosxsin²x}dx=[-1,1]∫cosxsin²xdx=[-1,1]∫sin²xd(sinx)
=(1/3)sin³x︱[-1,1]=(1/3)[sin³1-sin³(-1)]=(2/3)sin³1=(2/3)sin³(57°17′44.806″)=0.397215491...
设f(x)=e^(-x²)[in(x+1)/(1-x)],由于f(-x)=e^(-x²)[ln(-x+1)/(1+x)]=e^(-x²)[ln(x+1)/(1-x)]ֿ¹
=-e^(x²)[ln(x+1)/(1-x)]=-f(x),且x∈[-1,1],故f(x)是奇函数,∴[-1,1]∫e^(x²)[ln(x+1)/(1-x)]dx=0;
∴[-1,1]∫{e^(-x²)[in(x+1)/(1-x)]+cosxsin²x}dx=[-1,1]∫cosxsin²xdx=[-1,1]∫sin²xd(sinx)
=(1/3)sin³x︱[-1,1]=(1/3)[sin³1-sin³(-1)]=(2/3)sin³1=(2/3)sin³(57°17′44.806″)=0.397215491...
∫(-1,1)[e^(-x^2)[in(x+1)/(x-1)]+cosx(sinx)^2]dx=
∫(1+cosx)/(x+sinx)dx=?
∫[(1+sinx)/(1+cosx)]*(e^x)dx
∫sinx e^cosx dx不定积分 ∫(1/x^2)(sin(1/x))dx 不定积分
求一下两个不定积分:1.∫[xe^x/(e^x+1)^2]dx 2.∫dx/[(sinx)^3cosx]
∫(x+sinX)/(1+cosX)dx
∫[(x-cosx)/(1+sinx)]dx 不定积分,
1/(cosx+sinx)dx 积分怎样计算?还有e^-x^2dx积分谢谢
积分(1-cosx)dx/(x-sinx)
若∫f(x)dx=1/2x^2+C 则∫f(sinx)dx= -cosx+c
∫(2x-1)除以根号x dx ∫cosx dx +∫-2(sinx)^2 乘以cosx dx+∫(sinx)^4乘以c
求证 sin^2x/(sinx-cosx)-(sinx+cosx)/tan^2 x-1=sinx+cosx