Question:
Evaluate the following integrals:
$\int \frac{\sec x}{\log (\sec x+\tan x)} d x$
Solution:
Assume $\log (\sec x+\tan x)=t$
$\mathrm{d}(\log (\sec x+\tan x))=\mathrm{dt}$
(use chain rule to differentiate first differentiate $\log (\sec x+\tan x)$ then $(\sec x+\tan x)$
$\Rightarrow \frac{\sec x \tan x+\sec ^{2} x}{\sec x+\tan x} d x=d t$
$\Rightarrow \frac{\sec x(\tan x+\sec x)}{\sec x+\tan x} d x=d t$
$\Rightarrow \sec x d x=d t$
Put $\mathrm{t}$ and $\mathrm{dt}$ in the given equation we get
$\Rightarrow \int \frac{\mathrm{d} t}{t}$
$=\ln |t|+c$
But $t=\log (\sec x+\tan x)$
$=\ln |\log (\sec x+\tan x)|+c$
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