Dear Uncle Colin,

I need to figure out $\int \cos^3(2t) \sin^5(2t) \dt$ and I’m… just going round in circles. So to speak. What do you suggest?

- Doing Integration’s Really A Chore

Hi, DIRAC, and thanks for your message!

It’s very easy to end up going around in circles on these - the trick is to figure out the simplest transformation that turns your integral into something you *do* have the tools for.

I can see a fairly straightforward way to tackle this.

Products of powers of sines and cosines only really get *nice* when you wind up with a single sine or cosine multiplied by something horrible. In this case, writing the integral as $\int \cos(2t) \cos^2(2t) \sin^5(2t) \dt$ and replacing the cosine-squared with $1 - \sin^2(2t)$ gives:

$\int \cos(2t) \left[ \sin^5(2t) - \sin^7(2t)\right]\dt$

At first blush, that looks uglier - but both of the components are function-derivative and you end up with $\frac{1}{12}\sin^6(2t) - \frac{1}{16}\sin^8(2t) + C$.

Hope that helps!

- Uncle Colin

## Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008.
He lives with an espresso pot and nothing to prove.