Question

Need help on this calculus

Answer 1

`\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=\frac16u^6-\frac14u^4+C`

Back-substitute to get the antiderivative back in terms of
`t`:


`\frac16\cos^6t-\frac14\cos^4t+C`