Question

How do you evaluate the integral `int tan^5x`?

Answer 1

`int tan^5xdx = 1/4tan^4x - 1/2 tan^2x - ln abs (cosx) + C`

Explanation:

Using the trigonometric identity:

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

we can observe that in general

`int tan^nxdx = int (sec^2x-1)tan^(n-2)x dx = int tan^(n-2)xsec^2xdx-int tan^(n-2)x dx`

Remembering that:

`d/(dx) tan x = sec^2x`

the first integral is easy to solve:

`int tan^(n-2)x sec^2x dx = int tan^(n-2)x d(tanx) = 1/(n-1)tan^(n-1) +C`

So that we get the reduction formula:

`int tan^nxdx = 1/(n-1)tan^(n-1)x - int tan^(n-2)x dx`

Applying this formula twice we have:

`int tan^5xdx =1/4 tan^4x - int tan^3xdx = 1/4 tan^4x -1/2 tan^2x + int tanxdx`

We can solve this last integral directly:

`int tanx dx = int sinx/cosx dx = int (d(-cosx))/cosx = -ln abs (cosx) + C`

So that:

`int tan^5xdx = 1/4tan^4x - 1/2 tan^2x - ln abs (cosx) + C`