$\int \frac{t a n x}{s e c x + t a n x} d x$

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Solution

$I = \int \frac{tan x}{sec x + tan x} d x$
$= \int \frac{sin x}{1 + sin x} d x$
$= \int \frac{1 + sin x - 1}{1 + sin x} d x$
$= \int 1 - \frac{1}{1 + sin x} d x$
$= x - \int \frac{1 - sin x}{{cos}^{2} x} d x$
$= x - \int ({sec}^{2} x + sec x tan x) d x$
$x - tan x + sec x + c$ .

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