Trying to be Clever

This week I assigned my calculus class a problem* which involved the integral

\displaystyle\int \theta\sin\theta\cos\theta-\theta^2\cos^2\theta\ d\theta.

If you just dive in and do each term separately, this integral takes 3 applications of integration by parts (IBP). However, if you use your double angle identities, and do IBP on the second term first, then some terms gather up, and you only need one further application of IBP.

I think it is somewhat interesting, or fun, that the very similar integral

\displaystyle\int \theta\sin\theta\cos\theta+\theta^2\cos^2\theta\ d\theta

can be done with a single IBP. Perhaps it isn’t soo surprising, after you spend enough time trying to integrate these functions in various ways.

Now, in all honesty, the problem I assigned actually needed the definite integral

\displaystyle \int_0^{\pi} \theta\sin\theta\cos\theta-\theta^2\cos^2\theta\ d\theta.

I had thought maybe that since the integral is definite, you can use some other cleverness and not need IBP as many times. My idea was to shift the interval to [-\pi/2,\pi/2] using the substitution \alpha=\theta-\pi/2. Then we’ve got a symmetric interval, and odd functions will automatically drop out. Shifting sine or cosine by \pi/2 is not too inconvenient, because it just changes the function to the other (with sign changes in appropriate places).

After shifting, the odd functions drop out, and the remaining functions are even, so you can use symmetry yet again to do the integral over [0,\pi/2]. At some point you also change some \sin^2 \alpha‘s into \frac{1}{2}-\frac{1}{2}\cos 2\alpha‘s. When you think about \cos 2\alpha on the interval [0,\pi/2], it is odd-symmetric around the mid-point, \pi/4, so those terms drop out of the integral.

In the whole process so far, we have not used IBP at all. In fact, we’ve not yet done a single anti-derivative. Unfortunately, you’re stuck with a term \alpha^2\cos 2\alpha, and I can’t see how to get rid of it, so I break down and use IBP. Then you’ve still got an \alpha \sin 2\alpha, so IBP once more.

So I’ve done a lot of work to end up still doing IBP twice. I don’t know, probably that was all much more trouble than it was worth. But it was fun. Lots of nice applications of symmetry and some trig identities. I’d be delighted to hear any other solutions to this integral, so please share if you’ve got one.

* The problem I assigned: There is a cow tied to a circular silo with a piece of rope whose length is half the circumpherence of the silo. What is the area available to the cow for grazing? This is from Stewart’s Calculus textbook, in the section on parametric curves and areas.

Tags: ,

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

%d bloggers like this: