First, some other integrals:

`int "tan" x dx = int ("sin" x)/("cos" x) dx = `

`int (-du)/u = -"ln" |u| + C = - "ln" |"cos" x| + C`

also, we will need:

`"tan"^2 x = "sec"^2 x - 1`

`int "tan"^3 x dx = int "tan" x ("tan"^2 x) dx `

`= int "tan" x ("sec"^2 x -1) dx`

`= int "tan" x "sec"^2 x dx - int "tan" x dx`

The second integral we have already figure out

The first integral is a u substitution:

`u="tan" x` , `du = "sec"^2 x dx `

So:

`int "tan" x "sec"^2 x dx = int u du `

`= (1/2)u^2 + C = (1/2) "tan" ^2 x + C

Thus:

`int "tan"^3 x dx = (1/2) "tan" ^2 x + "ln" |"cos" x| + C

Now:

`int "tan"^5 x dx = int "tan"^3 x ("tan"^2 x) dx `

`= int "tan"^3 x ("sec"^2 x -1) dx`

`= int "tan"^3 x "sec"^2 x dx - int "tan"^3 x dx`

The second integral we have already figure out

The first integral is a u substitution:

`u="tan" x` , `du = "sec"^2 x dx `

So:

`int "tan"^3 x "sec"^2 x dx = int u^3 du `

`= (1/4)u^4 + C = (1/4) "tan" ^4 x + C

Thus:

`int "tan"^5 x dx = (1/4) "tan" ^4 x -(1/2) "tan" ^2 x - "ln" |"cos" x| + C `