Question

What's the integral of `int tan^5(x) dx`?

Answer 1

`1/4 sec^4x-sec^2x+ln abs secx`

Explanation:

`int tan^5x dx = int sin^5x/cos^5x dx`

`=int sin^4x(cosx)^-5 sinx dx`

`=int (1-cos^2x)^2(cosx)^-5 sinx dx`

`=int (1-2cos^2x + cos^4x)(cosx)^-5 sinx dx`

`=int ((cosx)^-5 -2(cosx)^-3 +(cosx)^-1) sinx dx`

Now integrate by substitution with `u = cosx`

Alternatively, one could use the reduction formula for `int tan^nx dx`. (If one has it memorized.)

Answer 2

Another method : )

`inttan^5(x)dx = inttan^2(x)*tan^3(x)dx =int (tan^2(x)+1-1)tan^3(x)dx`

`=int(tan^2+1)tan^3(x)dx-inttan^3(x)dx`

`=int(tan^2+1)tan^3(x)dx-int(tan^2(x)+1)tan(x)dx-inttan(x)dx` (By the same method)

For the two first integral let's `t=tan(x)`

So `dt = (1+tan^2(x))dx`

We have `intt^3dt - inttdt-inttan(x)dx`

`[1/4t^4]-[1/2t^2]+[ln(|cos(x)|)]+C`

(Don't forget absolute value, `ln(f(x))` can't take negative argument !)

Substitute back

`[1/4tan^4(x)]-[1/2tan^2(x)]+[ln(|cos(x)|)]+C`