`intsec^3tanx dx ` is the anti derivative of `sec^3xtanx` .

To integrate, use u-substitution method.

Let,

`u = sec x`

Differentiate u.

`du = sec x tan x dx`

Replacing the *x* variable in the above integral with *u* yields,

`intsec^3tanxdx = int (sec^2x)sec x tan x dx = intu^2du`

Use the power formula of integral which is `intu^n du = u^(n+1)/(n+1) +C` .

`= u^3/3 + C`

Then, substitute *u=sec x*.

`= (sec^3x)/3 + C`

**Hence, anti derivative of `sec^3xtanx` is `(sec^3x)/3 + C` .**

The integral `int sec^3 x tan x dx` has to be determined.

`int sec^3 x tan x dx`

=> `int 1/(cos^3 x)* (sin x/cos x) dx`

=> `int 1/(cos^4 x) * sin x dx`

let `cos x = y`

`(-1)*dy = sin x*dx`

=> `int (-1)*(1/y^4) dy`

=> `-1*y^(-3)/(-3) + C`

=> `y^(-3)/3 + C`

=> `1/(3*cos^3x) + C`

**The integral **`int sec^3 x tan x dx = 1/(3*cos^3x) + C`