Evaluate the integral.

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}si{n}^{3}6xdx$

Select the correct answer.

$\frac{1}{18}co{s}^{3}6x-\frac{1}{6}cos6x+C$

$\frac{1}{18}si{n}^{3}6x-\frac{1}{6}sin6x+C$

$\frac{1}{6}sin6x-\frac{1}{18}si{n}^{3}6x+C$

$\frac{1}{24}co{s}^{4}6x-\frac{1}{12}co{s}^{2}6x+C$

$\frac{1}{6}sin6x-\frac{1}{12}si{n}^{3}6x+C$

$\frac{1}{6}cos6x-\frac{1}{18}co{s}^{3}6x+C$

$\frac{1}{24}si{n}^{4}6x-\frac{1}{12}si{n}^{2}6x+C$