How do you integrate $(tan x) ln (cos x) d x$ $(tanx)ln(cosx)dx$ ?

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Answer 1 · 2016-08-01T13:30:48

$= - \frac{1}{2} {ln}^{2} cos x + C$ $= -1/2 ln^2 cos x + C$

well the first pattern to note is this

$\frac{d}{d x} (ln cos x) = \frac{1}{cos x} \cdot - sin x = - tan x$ $d/dx (ln cos x) = 1/ cos x * - sin x = - tan x$

And so

$\frac{d}{d x} ({ln}^{2} cos x) = 2 ln (cos x) \cdot \frac{d}{d x} (ln cos x)$ $d/dx (ln^2 cos x) = 2 ln (cos x) *d/dx (ln cos x)$

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

So final tweak: $\frac{d}{d x} (- \frac{1}{2} {ln}^{2} cos x) = ln (cos x) tan x$ $d/dx( -1/2 ln^2 cos x ) = ln (cos x) tan x$

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

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

$=- int \ u \ du$

$=- u^2/2 + C$

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

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