Question

Integrate log(cos x) dx .

Answer 1

Step-by-step explanation:

I hope it may help to you..

Answer 2

Answer:

$\tt{ \int {\log (\cos x)} \, dx = \dfrac{x\,\log (\cos x)}{2} + \dfrac{x\log |\sec x|}{2} + c }$

Step-by-step explanation:

Let $\tt{ I = \int {\log (\cos x)} \, dx }$

$\tt{ or \ I = \int {1\times\log (\cos x)} \, dx }$

w.k.t., $\tt{ \int {(I \times II)} \, dx = I\int {II} \, dx - \int{ [\ \{\ \dfrac{d}{dx}I\ \}\ \int {II} \, dx\ ] \, dx } + c }$

So, taking log(cos x) as first function and 1 as second function, we get

$\tt{ \ I = \log (\cos x)\int {1} \, dx - \int {[\ \{\ \dfrac{d}{dx}\log (\cos x)}\ \}\ \int {1} \, dx] \, dx + c }$

$\tt{ \implies \ I = x\,\log (\cos x) - \int {x\,\dfrac{(-\sin x)}{\cos x} \, dx + c }$

$\tt{ \implies \ I = x\,\log (\cos x) + \int {x\,\tan x \, dx + c }$

$\tt{ \implies \ I = x\,\log (\cos x) + x\int {\,\tan x \, dx + \int { [\ \{\ \dfrac{d}{dx}x \ \} \ \int {tanx} \, dx \ ] } \, dx + c }$

$\tt{ \implies \ I = x\,\log (\cos x) + x\log |\sec x| + \int { [\ 1 \times (-\log |\cos x|\,) \ ] } \, dx + c }$

$\tt{ \implies \ I = x\,\log (\cos x) + x\log |\sec x| - \int {log (\cos x)} \, dx + c }$

$\tt{ \implies \ I = x\,\log (\cos x) + x\log |\sec x| - I + c }$

$\tt{ \implies \ 2I = x\,\log (\cos x) + x\log |\sec x| + c }$

$\tt{ \implies \ I = \dfrac{x\,\log (\cos x)}{2} + \dfrac{x\log |\sec x|}{2} + c }$