First, to prevent loss in presentation marks, we always tag on the dx on the end of the integral. That is, our goal is to find int (tan^8 x) dx.
Notice that tan^8 x = tan^6 x tan^2 x = tan^6 x (sec^2 x -1) = tan^6 x sec^2 x - tan^6 x
Then int (tan^8 x) dx = int(tan^6 x)(sec^2 x)dx - int(tan^6 x)dx
To solve the first integral, we let u = tan x which implies that du = sec^2 x dx, thus the integral is simplified to
int u^6 du = u^7/7 + K = tan^7 x / 7 + K.
Using similar methods to find int(tan^6 x)dx, int(tan^4 x)dx and int(tan^2 x)dx, we derive that
int(tan^8 x)dx = tan^7 x / 7 - tan^5 x / 5 + tan^3 x / 3 - tan x + x + K
For a slightly shorter method, we use the method of reduction formulae. Consider the general integral I_n=int(tan^n x)dx. Using the same trigonometric identity, we find that
int (tan^n x) dx = int(tan^(n-2) x)(sec^2 x)dx - int(tan^(n-2) x)dx
which stands to reason that I_n = tan^(n-1) x / (n-1) - I_(n-2).
All we now need to do is find I_8, which in turn requires us to find I_6, I_4, I_2 and I_0 respectively.
I_0 is fairly easy, since int(tan^0 x) dx = int 1 dx = x + K.
Using the formulae given (I_n = tan^(n-1) x / (n-1) - I_(n-2)), we can find everything else.
I_2 = tan x - I_0 = tan x - x - K
I_4 = tan^3 x / 3 - I_2 = tan^3 x / 3 - tan x + x + K
I_6 = tan^5 x / 5 - I_4 = tan^5 x / 5 - tan^3 x / 3 + tan x - x - K
I_8 = tan^7 x / 7 - I_6 = tan^7 x / 7 - tan^5 x / 5 + tan^3 x / 3 - tan x + x + K
Therefore, int(tan^8 x)dx = tan^7 x / 7 - tan^5 x / 5 + tan^3 x / 3 - tan x + x + K