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Why does the value $-\ln(|\cos(x)|)$ become $\ln(|\sec(x)|)?$

I was doing an integral and I got my final answer as that, but I don't understand how you can just send the negative sign inside and make it $\sec(x).$

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    $\begingroup$ $\displaystyle{\large\cos\left(x\right) = {1 \over \sec\left(x\right)}}$. $\endgroup$ – Felix Marin Nov 11 '13 at 0:53
  • $\begingroup$ and $-\log(u) = \log(1/u)$, provided $u>0$. You may need some fast talking in case $\cos(x)=0$. $\endgroup$ – GEdgar Dec 27 '13 at 14:33
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You have $$a^b = c \iff b = \log_a(c). $$
Use this to show that for any power $r$, $\log_b(a^r) = r\log_b(a)$ for any base $b$.

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  • $\begingroup$ Oh my God, I didn't think of it like that and it confused me so much.. $\endgroup$ – Daniel Cook Nov 11 '13 at 0:48
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Because log(1/x) = -log(x), and sec = 1/cos

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Since $\cos x = \frac{1}{\sec x } $, then

$$ - \ln (\cos x ) = - \ln ( \frac{1}{\sec x} ) = - \ln 1 - (- \ln( \sec x )) = \ln(\sec x) $$

Where we have used the basic facts that

$\ln 1 = 0$

$\ln( \frac{a}{b}) = \ln a - \ln b$

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