Question

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

Answer 1

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

Explanation:

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

• `color(red)(tan^2 x = sec^2x - 1)`

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

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

`= 2x- tan x +C`