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