# Calculus: Hyperbolic Trigonometry, II

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

$\displaystyle\int\sqrt{1+x^2}\,dx.$

The usual technique involving circular trigonometric functions is to put $x=\tan(\theta),$ so that $dx=\sec^2(\theta)\,d\theta,$ and the integral transforms to

$\displaystyle\int\sec^3(\theta)\,d\theta.$

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

$\displaystyle\int\sec(\theta)\,d\theta.$

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

$\displaystyle\int \sec(\theta)\cdot\dfrac{\sec(\theta)+\tan(\theta)}{\sec(\theta)+\tan(\theta)}\,d\theta=\ldots$

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 $x=\sinh(u),$ so that $dx=\cosh(u)\,du$ and $\sqrt{1+x^2}=\cosh(u).$  Note well that taking the positive square root is always correct, since $\cosh(u)$ is always positive!

This results in the integral

$\displaystyle\int\cosh^2(u)\,du=\displaystyle\int\dfrac{1+\cosh(2u)}2\,du,$

which is quite simple to evaluate:

$\dfrac12u+\dfrac14\sinh(2u)+C.$

Now $u=\hbox{arcsinh}(x),$ and

$\sinh(2u)=2\sinh(u)\cosh(u)=2x\sqrt{1+x^2}.$

Recall from last week that we derived an explicit formula for $\hbox{arcsinh}(x),$ and so our integral finally becomes

$\dfrac12\left(\ln(x+\sqrt{1+x^2})+x\sqrt{1+x^2}\right)+C.$

You likely noticed that using a hyperbolic substitution is no more complicated than using the circular substitution $x=\sin(\theta).$  What this means is — no need to ever integrate

$\displaystyle\int\tan^m(\theta)\sec^n(\theta)\,d\theta$

again!  Frankly, I no longer teach integrals involving $\tan(\theta)$ and $\sec(\theta)$ 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

$\displaystyle\int\sqrt{x^2-1}\,dx.$

The usual technique?  Substitute $x=\sec(\theta),$ and transform the integral into

$\displaystyle\int\tan^2(\theta)\sec(\theta)\,d\theta.$

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

But not so using $x=\cosh(u).$  We get $dx=\sinh(u)\,du$ and $\sqrt{x^2-1}=\sinh(u)$ (here, a negative square root may be necessary).

We rewrite as

$\displaystyle\int\sinh^2(u)\,du=\displaystyle\int\dfrac{\cosh(2u)-1}2\,du.$

This results in

$\dfrac14\sinh(2u)-\dfrac u2+C=\dfrac12(\sinh(u)\cosh(u)-u)+C.$

All we need now is a formula for $\hbox{arccosh}(x),$ which may be found using the same technique we used last week for $\hbox{arcsinh}(x):$

$\hbox{arccosh}(x)=\ln(x+\sqrt{x^2-1}).$

Thus, our integral evaluates to

$\dfrac12(x\sqrt{x^2-1}-\ln(x+\sqrt{x^2-1}))+C.$

We remark that the integral

$\displaystyle\int\sqrt{1-x^2}\,dx$

is easily evaluated using the substitution $x=\sin(\theta).$  Thus, integrals of the forms $\sqrt{1+x^2},$ $\sqrt{x^2-1},$ and $\sqrt{1-x^2}$ may be computed by using the substitutions $x=\sinh(u),$ $x=\cosh(u),$ and $x=\sin(\theta),$ 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

$\displaystyle\int\sec(\theta)\,d\theta$

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 $x=\sinh(u)$ work above?  For the same reason $x=\tan(\theta)$ works: we can simplify $\sqrt{1+x^2}$ using one of the following two identities:

$1+\tan^2(\theta)=\sec^2(\theta)\ \hbox{ or }\ 1+\sinh^2(u)=\cosh^2(u).$

So $\sinh(u)$ is playing the role of $\tan(\theta),$ and $\cosh(u)$ is playing the role of $\sec(\theta).$  What does that suggest?  Try using the substitution $\sec(\theta)=\cosh(u)$!

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 $\sec(\theta)=\cosh(u),$ we have

$\sec(\theta)\tan(\theta)\,d\theta=\sinh(u)\,du.$

But notice that $\tan(\theta)=\sinh(u)$ — just look at the above identities and compare. We remark that if $\theta$ is restricted to the interval $(-\pi/2,\pi/2),$ then as a result of the asymptotic behavior, the substitution $\sec(\theta)=\cosh(u)$ gives a bijection between the graphs of $\sec(\theta)$ and $\cosh(u),$ and between the graphs of $\tan(\theta)$ and $\sinh(u).$ In this case, the signs are always correct — $\tan(\theta)$ and $\sinh(u)$ always have the same sign.

So this means that

$\sec(\theta)\,d\theta=du.$

What could be simpler?

Thus, our integral becomes

$\displaystyle\int\,du=u+C.$

But

$u=\hbox{arccosh}(\sec(\theta))=\ln(\sec(\theta)+\tan(\theta)).$

Thus,

$\displaystyle\int \sec(\theta)\,d\theta=\ln(\sec(\theta)+\tan(\theta))+C.$

Voila!

We note that if $\theta$ is restricted to the interval $(-\pi/2,\pi/2)$ as discussed above,  then we always have $\sec(\theta)+\tan(\theta)>0,$ 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 $\sec(\theta)$ 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….