What is #int sin(lnx) dx#?

We will proceed through integration by parts.

Let #I = intsin(ln(x))dx#

Integration by Parts (1)

Let #u = sin(ln(x))# and #dv = dx#

Then #du = cos(ln(x))/xdx# and #v = x#

Applying the integration by parts formula #intudv = uv - intvdu#

#I = intsin(ln(x))dx#

#= xsin(ln(x)) - intcos(ln(x))dx#

Integration by Parts (2)

Let #u = cos(ln(x))# and #dv = dx#

Then #du = -sin(ln(x))/x# and #v = x#

Applying the formula, we have

#intcos(ln(x))dx = xcos(ln(x)) - int-sin(ln(x))dx#

#= xcos(ln(x)) + intsin(ln(x))dx#

#= xcos(ln(x)) + I#

Substituting this into the result from the first integration by parts, we obtain

#I = xsin(ln(x)) - xcos(ln(x)) - I#

#=> 2I = x(sin(ln(x)) - cos(ln(x)))#

#=> I = (x(sin(ln(x)) - cos(ln(x))))/2#

But in the process of adding #I# to both sides, we lost the constant, and so for our final answer, we add it back in to get

#I = intsin(ln(x))dx = (x(sin(ln(x)) - cos(ln(x))))/2 + C#