[tex]\displaystyle\int\sin^3t\cos^3t\,\mathrm dt[/tex]
One thing you could do is to expand either a factor of [tex]\sin^2t[/tex] or [tex]\cos^2t[/tex], then expand the integrand. I'll do the first.
You have
[tex]\sin^2t=1-\cos^2t[/tex]
which means the integral is equivalent to
[tex]\displaystyle\int\sin t(1-\cos^2t)\cos^3t\,\mathrm dt[/tex]
Substitute [tex]u=\cos t[/tex], so that [tex]\mathrm du=-\sin t\,\mathrm dt[/tex]. This makes it so that the integral above can be rewritten in terms of [tex]u[/tex] as
[tex]\displaystyle-\int(1-u^2)u^3\,\mathrm du=\int(u^5-u^3)\,\mathrm du[/tex]
Now just use the power rule:
[tex]\displaystyle\int(u^5-u^3)\,\mathrm du=\dfrac16u^6-\dfrac14u^4+C[/tex]
Back-substitute to get the antiderivative back in terms of [tex]t[/tex]:
[tex]\dfrac16\cos^6t-\dfrac14\cos^4t+C[/tex]
