Question

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

Answer 1

First use a double-angle formula to replace `cos(2x)` by `2cos^{2}(x)-1`. Then distribute `cos(x)` through to rewrite your integrand as `(2cos^{3}(x)-cos(x))sin(x)`. Now do a substitution: `u=cos(x), du=-sin(x)dx`, making your integral transform to `\int(u-2u^{3})du=u^{2}/2-u^{4}/2+C=\frac{1}{2}\cos^{2}(x)-\frac{1}{2}\cos^{4}(x)+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 `-\frac{1}{16}\cos(4x)+C`.

Answer 2

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)/(n+1)` to integrate `x^n`.