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∫(-1,1)[e^(-x^2)[in(x+1)/(x-1)]+cosx(sinx)^2]dx=
人气:428 ℃ 时间:2020-05-20 08:40:08
解答
∫(-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...
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