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

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Answer 1 · 2016-08-01T13:30:48

#= -1/2 ln^2 cos x + C#

well the first pattern to note is this

#d/dx (ln cos x) = 1/ cos x * - sin x = - tan x#

And so

#d/dx (ln^2 cos x) = 2 ln (cos x) *d/dx (ln cos x) #

# = - 2 ln (cos x) tan x#

So final tweak: #d/dx( -1/2 ln^2 cos x ) = ln (cos x) tan x#

And thus

#int \ ln (cos x) tan x \ dx = int \ d/dx( -1/2 ln^2 cos x ) \ dx #

#= -1/2 ln^2 cos x + C#

Or you can try find a sub

This one works well:

#u = ln (cos x), du = - tan x \ dx#

...because the integral becomes

# int \tan x * u \* -1/tan x \ du#

# =- int \ u \ du#

# =- u^2/2 + C#

# =- 1/2 ln^2 (cos x) + C#