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

Question

Cameron Alexander · Accepted Answer

#= -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#

Axel Alvarez · Answer

undefined To integrate $ (\tan{x})\ln{(\cos{x})} $ with respect to $ x $, you can use integration by parts. Let $ u = \ln{(\cos{x})} $ and $ dv = \tan{x} \, dx $. Then, $ du = -\frac{\sin{x}}{\cos{x}} \, dx $ and $ v = -\ln{|\cos{x}|} $.

Applying the integration by parts formula, the integral becomes:

$$ \int (\tan{x})\ln{(\cos{x})} \, dx = -\ln{|\cos{x}|} \cdot \tan{x} + \int \ln{(\cos{x})} \cdot \frac{\sin{x}}{\cos{x}} \, dx $$

This second integral can be solved using substitution method. Let $ t = \cos{x} $, then $ dt = -\sin{x} \, dx $. Substituting into the integral, we get:

$$ \int \ln{(\cos{x})} \cdot \frac{\sin{x}}{\cos{x}} \, dx = -\int \ln{t} \, dt = -t\ln{t} + t + C $$

Substitute $ t = \cos{x} $ back in, we get:

$$ -\cos{x}\ln{|\cos{x}|} + \cos{x} + C $$

So, the final answer is:

$$ \int (\tan{x})\ln{(\cos{x})} \, dx = -\cos{x}\ln{|\cos{x}|} + \cos{x} + C $$