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 $\sec(x).$
$\endgroup$ 23 Answers
$\begingroup$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 $b$.
Because log(1/x) = -log(x), and sec = 1/cos
$\endgroup$ $\begingroup$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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