作业帮∫(-1,1)[e^(-x^2)[in(x+1)/(x-1)]+cosx(sinx)^2]dx=
来源：学生作业帮助网 编辑：作业帮 时间：2024/11/18 19:32:36
![∫(-1,1)[e^(-x^2)[in(x+1)/(x-1)]+cosx(sinx)^2]dx=](/uploads/image/z/8573094-54-4.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^(-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=
∫(-1,1)[e^(-x^2)[in(x+1)/(x-1)]+cosx(sinx)^2]dx=
∫(-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...
题目不对吧？在区间【－1 1】上，(x+1)/(x－1)<0，怎么能取对数呢？∫(-1,1)[e^(-x^2)[in(x+1)/(1-x)]+cosx(sinx)^2]dx= 是1-x的 ，我打错了e^(－x^2)是偶函数，ln(1+x)/(1－x)是奇函数，二者乘积是奇函数，积分值是0。 积分（cosx*(sinx)^2dx）＝(sinx)^3/3|上限1下限－1＝(2sin1)/3...
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题目不对吧？在区间【－1 1】上，(x+1)/(x－1)<0，怎么能取对数呢？
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