Calculus: Hyperbolic Trigonometry, II

Now on to some calculus involving hyperbolic trigonometry!  Today, we’ll look at trigonometric substitutions involving hyperbolic functions.

Let’s start with a typical example:

The usual technique involving circular trigonometric functions is to put so that and the integral transforms to

In general, we note that when taking square roots, a negative sign is sometimes needed if the limits of the integral demand it.

This integral requires integration by parts, and ultimately evaluating the integral

And how is this done?  I shudder when calculus textbooks write

How does one motivate that “trick” to aspiring calculus students?  Of course the textbooks never do.

Now let’s see how to approach the original integral using a hyperbolic substitution.  We substitute so that and  Note well that taking the positive square root is always correct, since is always positive!

This results in the integral

which is quite simple to evaluate:

Now and

Recall from last week that we derived an explicit formula for and so our integral finally becomes

You likely noticed that using a hyperbolic substitution is no more complicated than using the circular substitution  What this means is — no need to ever integrate

again!  Frankly, I no longer teach integrals involving and which involve integration by parts.  Simply put, it is not a good use of time.  I think it is far better to introduce students to hyperbolic trigonometric substitution.

Now let’s take a look at the integral

The usual technique?  Substitute and transform the integral into

Sigh.  Those irksome tangents and secants.  A messy integration by parts again.

But not so using  We get and (here, a negative square root may be necessary).

We rewrite as

This results in

All we need now is a formula for which may be found using the same technique we used last week for

Thus, our integral evaluates to

We remark that the integral

is easily evaluated using the substitution  Thus, integrals of the forms and may be computed by using the substitutions and respectively.  It bears repeating:  no more integrals involving powers of tangents and secants!

One of the neatest applications of hyperbolic trigonometric substitution is using it to find

without resorting to a completely unmotivated trick.  Yes, I saved the best for last….

So how do we proceed?  Let’s think by analogy.  Why did the substitution work above?  For the same reason works: we can simplify using one of the following two identities:

So is playing the role of and is playing the role of  What does that suggest?  Try using the substitution !

No, it’s not the first think you’d think of, but it makes sense.  Comparing the use of circular and hyperbolic trigonometric substitutions, the analogy is fairly straightforward, in my opinion.  There’s much more motivation here than in calculus textbooks.

So with  we have

But notice that — just look at the above identities and compare. We remark that if is restricted to the interval then as a result of the asymptotic behavior, the substitution gives a bijection between the graphs of and and between the graphs of and In this case, the signs are always correct — and always have the same sign.

So this means that

What could be simpler?

Thus, our integral becomes

But

Thus,

Voila!

We note that if is restricted to the interval as discussed above,  then we always have so there is no need to put the argument of the logarithm in absolute values.

Well, I’ve done my best to convince you of the wonder of hyperbolic trigonometric substitutions!  If integrating didn’t do it, well, that’s the best I’ve got.

The next installment of hyperbolic trigonometry?  The Gudermannian function!  What’s that, you ask?  You’ll have to wait until next time — or I suppose you can just google it….