∫(-1,1){e^(-x²)[in(x+1)/(1-x)]+cosxsin²x}dx
設(shè)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)是奇函數(shù),∴[-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...