Find $\int \ln(\tan(x))/(\sin(x) \cos(x))dx$

Question

I was given this question in a review package, and it has me stumped:

I started off using the identity $\tan(x) = \sin(x) / \cos(x)$ and then used the fact that $\sin(x) \cos(x) = .5\sin(2x)$ to try and simplify the denominator. I looked around for a basic $u$ substitution but couldn't find any. I broke the $ln(\sin(x)/\cos(x))$ into $\ln(\sin(x)) - ln(cos(x))$ thinking I could maybe split the integral in two and that would help, but to no avail. Using parts looks very messy and I'm pretty lost at this point, anyone know how to get me on the right track?

Answer 1

Let $u = \ln(\tan x)$. It's simpler than you think.

Answer 2

Put $\ln \tan x =t$

therefore $${\sec^2x \over \tan x} \, \mathrm dx=\mathrm dt$$

$$\mathrm dx={\mathrm dt \, \sin x \cos x}$$

Subsitute these values into the integral

$$I=\int t \, \mathrm dt$$

$$I={t^2 \over 2}+constant$$

$$I={(\ln (\tan x))^2 \over 2}+constant$$

Answer 3

\begin{eqnarray} \int\frac{\ln\tan x}{\sin x\cos x}dx&=&\int\frac{\ln\tan x}{\tan x}\sec^2xdx =\int\frac{\ln\tan x}{\tan x}d\tan x\\ &=&\int\ln\tan xd\ln\tan x=\frac{1}{2}(\ln\tan x)^2+C \end{eqnarray}

Answer 4

We know that $\sin x \cos x = \frac{1}{2}\sin 2x$

We know that $\tan x = \csc 2x - \cot 2x$

So

$$ \int\frac{\ln\tan x}{\sin x\cos x}dx=\int\frac{2\ln(\csc(2x) - \cot(2x))}{\sin2x}dx\\ =\int2\ln(\csc(2x) - \cot(2x))\csc(2x)dx\\ $$

Do you know the integral of $\csc x$? It's $\ln(\csc x - \cot x)$

I think that's enough for now...