# How do you integrate (sinx)(cosx)(cos2x)dx?

Mar 15, 2015

First use a double-angle formula to replace $\cos \left(2 x\right)$ by $2 {\cos}^{2} \left(x\right) - 1$. Then distribute $\cos \left(x\right)$ through to rewrite your integrand as $\left(2 {\cos}^{3} \left(x\right) - \cos \left(x\right)\right) \sin \left(x\right)$. Now do a substitution: $u = \cos \left(x\right) , \mathrm{du} = - \sin \left(x\right) \mathrm{dx}$, making your integral transform to $\setminus \int \left(u - 2 {u}^{3}\right) \mathrm{du} = {u}^{2} / 2 - {u}^{4} / 2 + C = \setminus \frac{1}{2} \setminus {\cos}^{2} \left(x\right) - \setminus \frac{1}{2} \setminus {\cos}^{4} \left(x\right) + C .$ There are lots of alternative ways of writing this answer because of all the trigonometric identities out there. You could check, for instance, that it is equivalent to $- \setminus \frac{1}{16} \setminus \cos \left(4 x\right) + C$.

Mar 15, 2015

You can try this:

where at the end you treat $\sin$ and $\cos$ as $x$ in a normal integral where you would use the form: ${x}^{n + 1} / \left(n + 1\right)$ to integrate ${x}^{n}$.