How do you integrate #cos(log(x)) dx#?

1 Answer
May 14, 2015

(I will assume #logx# is natural log. (If not, insert ln10 where needed.)

#int cos(log(x))dx = int xcos(log(x))*1/xdx#

Let #u=x# and #dv# is the rest of the integrand.
Now we can integrate #v = int cos(log(x))*1/xdx = sin(log(x))#
(Use substitution with #w=log(x)#)

Parts gives us:

#int cos(log(x))dx = xsin(log(x)) - int sin(log(x)) dx#

Do the same trick again to get

#int cos(log(x))dx = xsin(log(x)) + xcos(log(x)) - int cos(log(x)) dx#

So

#int cos(log(x))dx = 1/2(xsin(log(x)) + xcos(log(x))) #