# What is the integral `int sec^2x*tan x dx` . Does this require breaking down the trigonometric terms.

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### 1 Answer

The integral `int sec^2x*tan x dx` has to be determined.

This can be done with the substitution `y = tan x => dy = sec^2x dx`

`int sec^2x*tan x dx`

=> `int y dy`

=> `y^2/2 + 2`

replace y with tan x

=> `(tan^2x) /2 + C`

You could write `sec^2x*tan x` = `(1/cos^2x)*(sin x/cosx)` = `sin x/(cos^3x)` and then substitute `cos x = y => dy = -sin x dx` . This gives

`intsec^2x*tan x dx `

=` int sin x/(cos^3x) dx`

= `-int 1/y^3 dy`

= `y^-2/2`

= `1/(2*cos^2x)`

This is equivalent to `(tan^2x)/2`

**The integral **`int sec^2x*tan x dx = (tan^2x)/2 + C = 1/(2*cos^2x) + C`