- #1

Knissp

- 75

- 0

## Homework Statement

[tex]\int tan^5(3x) sec^2(3x) dx [/tex]

## The Attempt at a Solution

[tex]u = tan(3x)[/tex]

[tex]du = 3 sec^2(3x) dx [/tex]

[tex]du/3 = sec^2(3x) dx [/tex]

[tex]\int tan^5(3x) sec^2(3x) dx [/tex]

[tex]= \int 1/3 u^5 du[/tex]

[tex]= u^6/18 + C[/tex]

[tex]= tan^6(3x)/18 + C [/tex]

The calculator says the integral should be:

[tex]\frac {1 - 3 sin^2(3x) cos^2(3x)}{18cos^6(3x)} + C [/tex]

The answer I got does not differ from the calculator by a constant. Any help would be appreciated.