Question

How do you integrate `tan(x)^3`?

Answer 1

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

Explanation:

You can write `tan^3x` as:

`tan^3x = (sin^3x)/(cos^3x) = sinx (1-cos^2x)/(cos^3x)`

So that:

`int tan^3x dx= int sinx (1-cos^2x)/(cos^3x)dx`

Now substitute:

`t=cosx`
`dt = -sint dt`

and you have:

`int tan^3x dx= -int (1-t^2)/(t^3)dt`

Separating the sum:

`int tan^3x dx= -int 1/(t^3)dt + int (dt)/t = 1/(2t^2)+ln abs(t)`

and substituting back `t=cosx`

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