# Evaluate the indefinite integral integrate of tan^6(x)dx

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### 1 Answer

You should use the following substitution such that:

`tan x = t => 1/(cos^2 x) dx = dt`

`tan^6 x = tan^4 x*tan^2 x => tan^6 x = tan^4 x*(sin^2 x)/(cos^2 x)`

You need to integrate both sides such that:

`int tan^6 x dx = int tan^4 x*(sin^2 x)/(cos^2 x) dx`

`int tan^6 x dx = int tan^4 x*(1 - cos^2 x)/(cos^2 x) dx`

Using the linearity of integrals yields:

`int tan^6 x dx = int tan^4 x*(1/(cos^2 x)) dx - int tan^4 x dx`

`int tan^4 x*(1/(cos^2 x)) dx = int t^4 dt = t^5/5 + c`

Substituting back `tan x` for t yields:

`int tan^4 x*(1/(cos^2 x)) dx = (tan^5 x)/5 + c`

`int tan^4 x dx = int tan^2 x*(1-cos^2 x)/(cos^2 x)dx`

`int tan^4 x dx = int tan^2 x*(1/(cos^2 x)) dx - int tan^2 x dx`

`int tan^2 x*(1/(cos^2 x)) dx = int t^2 dt = t^3/3 + c`

Substituting back `tan x` for t yields:

`int tan^2 x*(1/(cos^2 x)) dx = (tan^3 x)/3 + c`

`int tan^2 x dx = int (1 - cos^2 x)/(cos^2 x)dx`

`int tan^2 x dx = int 1/(cos^2 x) dx - int dx`

`int tan^2 x dx = tan x - x + c`

`int tan^4 x dx = (tan^3 x)/3 - tan x + x + c`

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

**Hence, evaluating the given integral under the given conditions yields `int tan^6 x dx = (tan^5 x)/5 - (tan^3 x)/3+ tan x - x + c.` **