f (x) = ∫[a sin(ln x) + b cos(ln x)]dx
1个回答

设lnx=y则x=e^y

asin(lnx)dx积分=asinyd(e^y)=asiny*e^y-ae^yd(siny)=asiny*e^y-acosyd(e^y)

asiny*e^y-acosy*e^y+ae^yd(cosy)=a*e^y(siny-cosy)-asiny*e^ydy

2asiny*d(e^y)=a*e^y(siny-cosy)

asiny=a*e^y(siny-cosy)/2

asin(lnx)dx积分=a*x{sin(lnx)-cos(lnx)}/2

同理bcos(lnx)dx积分=b*x{sin(lnx)+cos(lnx)}/2

f(x)=[ax{sin(lnx)-cos(lnx)}+bx{sin(lnx)+cos(lnx)}]/2

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