# Calculus Problem of the Week August 12, 2011

Sometimes easy and sometimes hard, our calculus problem of the week could come from any calculus topic. If you really want to get better at calculus, following these problems is a great way to make yourself practice! Past calculus problems of the week.

This week’s problem:
(click “see the solution” at the bottom of post to, well, see the solution.)

Find the integral: $\int(\tan(x)\tan(2x)\tan(3x))dx$. Hint: $\tan(3x)=\tan(2x+x)$.

See the solution.

I gave you the hint above to try to get you to think about using some identities here as trying this integral directly could be insane!

Using the identity for sums of Tangent, we know that $\tan(3x)=\tan(2x+x)=\dfrac{\tan(2x)+\tan(x)}{1-\tan(x)\tan(2x)}$. But, if you simply replace the term in the integral with this, you are STILL making things more complicated than they need be. Let’s take a closer look at this identity and see what other information is in it.

$\tan(3x)=\dfrac{\tan(2x)+\tan(x)}{1-\tan(x)\tan(2x)}$

If we cross multiply:

$\tan(3x)(1-\tan(x)\tan(2x))=\tan(2x)+\tan(x)$

then distribute:

$\tan(3x)-\tan(x)\tan(2x)\tan(3x)=\tan(2x)+\tan(x)$

Aha – do you see that our integral is right in there? If you solve for $\tan(x)\tan(2x)\tan(3x)$ in this identity then you find that $\tan(x)\tan(2x)\tan(3x)=\tan(3x)-\tan(2x)-\tan(x)$. Why is this such a big deal? It looks really similar to what we had before right? Well, you can’t just take the integral of individual terms when they are multiplied but you can if the terms are added or subtracted. In other words this means that:

$\int(\tan(x)\tan(2x)\tan(3x))dx=\int(\tan(3x)-\tan(2x)-\tan(x))dx$
$=\int(\tan(3x))dx-\int(\tan(2x))dx-\int(\tan(x))dx$

Now you can use u substitution ($u=3x,2x$ respectively) and the fact that $\int(\tan(x))=-\ln|\cos(x)|$ to find the integral.

$\int(\tan(x)\tan(2x)\tan(3x))dx=\int(\tan(3x)-tan(2x)-tan(x))dx$
$=-\dfrac{1}{3}\ln|\cos(3x)|+\dfrac{1}{2}\ln|\cos(2x)|-\ln|\cos(x)|+C$