How do you evaluate the integral of #int tan(x)ln(cosx) dx#?

1 Answer
May 3, 2016

#-ln^2(cosx)/2+C#

Explanation:

We can solve this integral using u-substitution, but it is fairly unapparent from the outset.

If we let #u=ln(cosx)#, differentiating this reveals that

#du=(d/dx(cosx))/cosxdx=(-sinx)/cosxdx=-tanxdx#

Multiply the interior and exterior of the integral by #-1#.

#inttanxln(cosx)dx=-int-tanxln(cosx)dx=-intudu#

Integrating this gives #-u^2/2+C#, and since #u=ln(cosx)#, this becomes #-ln^2(cosx)/2+C#.