How do you integrate #int sin(lnx)# using integration by parts?

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Answer 1 · 2015-11-03T22:07:46

#1/2([xsin(ln(x))]-[xcos(ln(x))])#

#intsin(ln(x))dx#

Let's #u = ln(x)#

#du = dx/x#

#dx=xdu#

#x=e^u#

then

#inte^usin(u)du#

By part : #dv = e^u ; v = e^u ; w = sin(u) ; dw = cos(u) #

#[v*w]-intdw*v#

#inte^usin(u)du=[e^usin(u)]-inte^ucos(u)du#

By part again

#dv=e^u ; v = e^u ; w = cos(u) ; dw = -sin(u)#

#inte^usin(u)du=[e^usin(u)]-([e^ucos(u)]+inte^usin(u)du)#

#inte^usin(u)du=[e^usin(u)]-[e^ucos(u)]-inte^usin(u)du#

#2inte^usin(u)du=[e^usin(u)]-[e^ucos(u)]#

#inte^usin(u)du=1/2([e^usin(u)]-[e^ucos(u)])#

Substitute back for #u = ln(x)#

#intsin(ln(x))dx=1/2([xsin(ln(x))]-[xcos(ln(x))])#