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Thread: Integration

  1. #1
    Junior Member
    Joined
    Jan 2009
    Posts
    74

    Integration

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

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

    to...

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

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

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

    I appreciate the help! Thank you!
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  2. #2
    Senior Member
    Joined
    Dec 2007
    From
    Anchorage, AK
    Posts
    276
    The first (outside) minus sign in $\displaystyle y=-[-\frac{\ln(2)}{2}-0]$ is the minus sign from $\displaystyle y=-\left.\frac{\ln|\cos(2x)|}{2}\right|_0^{\frac{\pi} {6}}$; note there is also a minus sign on $\displaystyle -\frac{\ln(2)}{2}$; this is because $\displaystyle \ln\left(\frac{1}{2}\right)=\ln\left(2^{-1}\right)=-\ln{2}$, from the basic properties of logarithms (which you should know).

    --Kevin C.
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