Integration

**juicysharpie** · March 14th 2009, 01:34 PM

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!

**TwistedOne151** · March 14th 2009, 01:52 PM

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.

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