How do you find $int 1-tanx^2dx$ ?

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Answer 1 · 2015-12-03T01:18:31

$int(1-tan^2x)dx = 2x- tan x +C$

We can start of by doing the following

$int 1-tan^2x dx = int 1 dx - int tan^2x dx$

Then integrate the first integral

$= x- int(sec^2x -1)dx$

Rewrite using trigonometric identity

$= x- [int(sec^2x) dx -int1dx]$
$color(red)(int (sec^2x dx= tan x + C)$

$= x- [ tan x -x] +C$

$= 2x- tan x +C$

How do you find int 1-tanx^2dxint 1-tanx^2dx?