# Evaluate the indefinite integral integrate of (tan(6x))^3(sec(6x))dx

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### 1 Answer

You should remember that `sec 6x = 1/(cos 6x)` and `tan 6x = (sin 6x)/(cos 6x)` such that:

`int tan^3(6x)sec 6x dx = int ((sin^3 6x)/(cos^3 6x))*(1/(cos 6x)) dx`

`int tan^3(6x)sec 6x dx = int (sin^2 6x)/(cos^2 6x)*(1/(cos^2 6x)) dx`

`int tan^3(6x)sec 6x dx = int tan^2 6x*(1/(cos^2 6x)) dx`

You should come up with the following substitution such that:

`tan 6x = t => 6/(cos^2 6x) dx = dt => 1/(cos^2 6x) dx = (dt)/6`

Changing the variable yields:

`int tan^2 6x*(1/(cos^2 6x)) dx = int t^2*(dt)/6`

`int t^2*(dt)/6 = (1/6)(t^3/3) + c`

Substituting back `tan 6x` for t yields:

`int tan^3(6x)sec 6x dx = (tan^3 6x)/18 + c`

**Hence, evaluating the indefinite integral yields `int tan^3(6x)sec 6x dx = (tan^3 6x)/18 + c.` **