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#
