1. ## Integration

My book did something confusing. I'm not sure how to get from one step to the other.

Step 1: $y = - \left. \frac{ln|cos(2x)|}{2}\right| _0^\frac{\pi}{6}$

to...

Step 2: $y = -[-\frac{ln(2)}{2} - 0]$

But I got $cos 2x$ when $\frac{\pi}{6}$ is $\frac{\pi}{3}$ and $cos \frac{\pi}{3} = \frac{1}{2}.$
How did it turn into 2 instead of 1/2? And also, where did the $-$ sign in front of the whole thing come from?

The original problem is to evaluate: $y = \int_0^ \frac{\pi}{6} tan(2x)dx$

I appreciate the help! Thank you!

2. The first (outside) minus sign in $y=-[-\frac{\ln(2)}{2}-0]$ is the minus sign from $y=-\left.\frac{\ln|\cos(2x)|}{2}\right|_0^{\frac{\pi} {6}}$; note there is also a minus sign on $-\frac{\ln(2)}{2}$; this is because $\ln\left(\frac{1}{2}\right)=\ln\left(2^{-1}\right)=-\ln{2}$, from the basic properties of logarithms (which you should know).

--Kevin C.