Answer
$\frac{tan^53x}{15}- \frac{tan^33x}{9} + \frac{tan3x}{3}- x+C$
Work Step by Step
Find the indefinite integral
$\int tan^6(3x)dx$
Rewrite the integrand to simplify things, Let 3x=x
$\frac{1}{3} \int tan^6xdx$
Use Guideline #3
$\frac{1}{3} \int tan^6xdx = \frac{1}{3}\int tan^4x(sec^2x -1)dx$
$\frac{1}{3}[\int tan^4x sec^2x dx - \int tan^4xdx]$
$\frac{1}{3}[\int tan^4xsec^2xdx - \int tan^2x(sec^2x -1) dx]$
Let $u=tanx$ then $du=sec^2xdx$
$\frac{1}{3}[ \int u^4du - \int u^2du + \int tan^2xdx]$
$\frac{1}{3}[\int(u^4 - u^2)du + \int(sec^2x -1)dx]$
$\frac{1}{3}[\frac{u^5}{5}- \frac{u^3}{3}] + \frac{1}{3}[tanx - x] +C$
$\frac{tan^5x}{15}- \frac{tan^3x}{9} + \frac{tanx}{3}- \frac{x}{3}+C$
Resubstitute $x=3x$
$\frac{tan^53x}{15}- \frac{tan^33x}{9} + \frac{tan3x}{3}- x+C$
