Before we evaluate this integral, I'd like to say that this problem might be very lengthy.
Also, let me note some strategies for evaluating #tan x# integrals:
Trigonometric identity:
#tan^2(x) = sec^2(x)-1#
Power reduction formula:
#int tan^n(x) dx = int tan^(n-2)(x) * (sec^(2)(x)-1) dx #
We're going to use the same kind of strategies when evaluating this integral, thus giving us
#int tan^6(x) dx = int tan^4(x)(sec^2(x)-1) dx#
#= int underbrace(tan^4(x))_(u^4)* underbrace(sec^2(x))_(du)- tan^4(x) dx#
Note: Let #u = tan x -> du = sec^2(x) dx#
Substitute and integrate to get #int u^4 du = (u^5)/5 + C = (tan^5(x))/5#
#= (tan^5(x))/5 - int tan^4(x) dx#
#= (tan^5(x))/5 - int tan^2(x)(sec^2(x)-1) dx#
#=(tan^5(x))/5 - int underbrace(tan^2(x))_(u^2)* underbrace(sec^2(x))_(du)-tan^2(x) dx#
Note: Let #u = tan x -> du = sec^2(x) dx#
Substitute and integrate to get #int u^2 du = (u^3)/3 + C = (tan^3(x))/3#
#= (tan^5(x))/5-(tan^3(x))/3 + int tan^2(x) dx#
#=(tan^5(x))/5-(tan^3(x))/3 + int (sec^2(x)-1) dx#
Note that #int sec^2(x) dx = tan x + C#
#=(tan^5(x))/5-(tan^3(x))/3 + tan x - x + C#
