Question

How do you integrate `(tanx)ln(cosx)dx`?

Answer 1

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

Explanation:

Lets choose: `u=ln(cosx)`, then:
`du=(ln(cosx))'dx=(1/cosx)(cosx)'dx=(1/cosx)(-sinx)dx=-sinx/cosxdx=-tanxdx`
So, `tanxdx=-du` and
`inttanxln(cosx)dx=int(-du)u=-u^2/2+C=-ln^2(cosx)/2+C`