# Evaluate the indefinite integral integrate of (sec^2(t))/(sqrt(1-9tan^2(t)))dt

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### 1 Answer

You should come up with the following substitution such that:

`tan t = u => 1/(cos^2t) dt = du => sec^2 t dt = du`

Changing the variable yields:

`int (sec^2 t)/(sqrt(1 - 9tan^2 t))dt = int (du)/(sqrt(1-9u^2))`

You should come up with the following substitution such that:

`u = (sin v)/3 => u^2 = (sin^2 v)/9`

`du = (cos v)/3 dv`

Changing the variable yields:

`int (du)/(sqrt(1-9u^2)) = int (cos v)/3/(sqrt(1-9((sin^2 v)/9))) dv`

`(1/3) int cos v/(sqrt(1 - sin^2 v)) dv`

You need to use the fundamental formula of trigonometry such that:

`1 - sin^2 v = cos^2 v => sqrt(1 - sin^2 v) = sqrt(cos^2 v) = cos v`

`(1/3) int cos v/(sqrt(1 - sin^2 v)) dv = (1/3) int cos v/cos v dv`

`(1/3) int cos v/(sqrt(1 - sin^2 v)) dv = (1/3) int dv`

`(1/3) int cos v/(sqrt(1 - sin^2 v)) dv = (1/3) v + c`

Substituting back `arcsin (3u)` fot v yields:

`int (du)/(sqrt(1-9u^2)) = (1/3)arcsin (3u) + c`

Substituting back `tan t` for u yields:

`int (sec^2 t)/(sqrt(1 - 9tan^2 t))dt = (1/3) arcsin (3tan t) + c`

**Hence, evaluating the given integral yields `int (sec^2 t)/(sqrt(1 - 9tan^2 t))dt = (1/3) arcsin (3tan t) + c.` **