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#
