What is the integral of int (tan x)^6dx?

1 Answer
Jul 13, 2016

int tan^6(x) dx =(tan^5(x))/5-(tan^3(x))/3 + tan x - x + C

Explanation:

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