Question
$\int \sec ^2 x d x$

Answer

$\text { Let } I =\int \sec ^2 x d x$
$=\int \sec x \cdot \sec ^2 x d x$
$=\sec x \int \sec ^2 x d x-\int\left[\frac{ d }{ d x}(\sec x) \int \sec ^2 x d x\right] d x$
$=\sec x \cdot \tan x-\int \sec x \tan x \cdot \tan x d x$
$=\sec x \cdot \tan x-\int \sec x \tan { }^2 x d x$
$=\sec x \cdot \tan x-\int \sec x\left(\sec { }^2 x-1\right) d x$
$=\sec x \cdot \tan x-\int \sec { }^3 x d x+\int \sec x d x$
$\therefore \text { I }=\sec x \cdot \tan x- I +\log |\sec x+\tan x|+ c _1$
$\therefore \log [\sec x \tan x+\log |\sec x+\tan x|]+ c \text { where c }=\frac{ c _1}{2}$

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