Integral of $\ln(\cos x) \tan(x)$

Question

How can I find $$\int \ln(\cos x)\tan(x) dx$$

I'm having a hard time all the site I usually use to check doesn't work for this for some reason...

I got $$\frac{\ln(\cos x)}{2}$$

Answer 1

Hint: Take $ u =\cos x$ then $du = -\sin x dx$ then

$$-\int \frac{\ln u}{u} du = -\frac{\ln^2 u}{2} + C$$

Answer 2

Just set $u=\ln(\cos(x))$, $du=\frac{-\sin(x)}{\cos(x)}dx=-\tan(x)dx$

Thus, the required integral is

$I=\int u\cdot\tan(x)\cdot\frac{du}{-\tan(x)}=\int -udu=\frac{-u^2}{2}+c=-\frac{\ln(cos(x))^2}{2}+c$

Answer 3

let $$u=\ln(\cos(x))$$ $$du=\frac{-\sin(x)}{\cos(x)}dx=-\tan(x)dx$$ $$\int u(-du)=\int-udu$$

Answer 4

One way of thinking about substitutions is $$\int f(u(x))u'(x)dx = \int f(u)du\text{ , where }du=u'(x)dx$$ An alternative view is $$\int f(u(x))u'(x)dx = \int f(u(x))u'(x)dx\frac{\frac{du}{dx}}{u'(x)}$$

In particular $$\int \ln(\cos x)\tan x dx=\int \ln(\cos x)\tan x dx\cdot\frac{\frac{d\cos x}{dx}}{\frac{d\cos x}{dx}}=\int\frac{\ln (\cos x)\tan x}{-\sin x}d\cos x$$ The benefit of this approach is that it visually reminds me that I need to divide by the derivative of the subsitution (so I can easily see what is going to disappear). This eliminates the guesswork of robotically applying a $du=\dots dx$ procedure.

Continuing $$\int\frac{\ln (\cos x)\sin x}{-\sin x \cos x}d\cos x = \int\frac{\ln (\cos x)}{- \cos x}d\cos x=\int\frac{\ln u}{-u}du=-\int\frac{\frac{1}{u}\ln u}{1}du=$$

What is $\frac{d}{du}\ln u$?