How do you integrate #int cos(lnx)# by integration by parts method?

This isn't necessarily done by only integration by parts, but it's one step in the process; this actually looks like one where you simply choose your methods and see what works.

#int cos(lnx)dx = 1/2 x(sin(lnx) + cos(lnx)) + C#

First, let us try:

#u = lnx#
#du = 1/xdx#

Then, we have:

#int cos(u)xdu = int e^ucosudu#

This looks more reasonable, and actually would be done using integration by parts, since we have a function where #(df)/(dx) = f(x)# (#e^u)#, and a function with cyclic derivatives (#cosu#).

Let:

#s = e^u#
#dt = cosudu#
#t = sinu#
#ds = e^udu#

This gives:

#int sdt = st - int tds#

#= e^usinu + int e^u(-sinu)du#

Since #sin# derivatives are cyclic, repeat the process. Let:

#s = e^u#
#dt = -sinudu#
#t = cosu#
#ds = e^udu#

Therefore:

#int e^ucosudu = int sdt#

#= e^usinu + [e^ucosu - int e^ucosudu]#

If you notice, the original integral has shown up again, so:

#2int e^ucosudu = e^usinu + e^ucosu = e^u(sinu + cosu)#

#=> int e^ucosudu = 1/2e^u(sinu + cosu)#

And now, we undo the substitution, using #u = lnx#, to get:

#int cos(lnx)dx = color(blue)(1/2 x(sin(lnx) + cos(lnx)) + C)#