Old school integral

This isn’t going to be one of my typical posts, but I just cracked a challenging indefinite integral and wanted to share.

I made a mistake solving a calculus problem a few weeks ago and ended up at an integral that looked pretty simple.  I tried several approaches and found many dead ends before finally getting a breakthrough.  Rather than just giving a proof, I thought I’d share my thought process in hopes that some students just learning integration techniques might see some different ways to attack a problem and learn to persevere through difficult times.

In my opinion, most students taking a calculus class would never encounter this problem.  The work that follows is clear evidence why everyone doing math should have access to CAS (or tables of integrals when CAS aren’t available).

Here’s the problem:

Integrate $\int \left( x^2 \cdot \sqrt{1+x^2} \right) dx$.

For convenience, I’m going to ignore in this post the random constant that appears with indefinite integrals.

While there’s no single algebraic technique that will work for all integrals, sometimes there are clues to suggest productive approaches.  In this case, the square root of a binomial involving a constant and a squared variable term suggests a trig substitution.

From trig identities, I knew $tan^2 \theta + 1 = sec^2 \theta$, so my first attempt was to let $x=tan \theta$, which gives $dx=sec^2 \theta d\theta$.  Substituting these leads to

$(tan \theta)'=sec^2 \theta$, claiming two secants for the differential in a reversed chain rule, but left a single secant in the expression, so I couldn’t make the trig identities work because odd numbers of trigs don’t convert easily using Pythagorean identities.  Then I tried using $(sec \theta)'=sec \theta \cdot tan \theta$, leaving a single tangent after accounting for the potential differential–the same problem as before.  A straightforward trig identity wasn’t going to do the trick.

Then I recognized that the derivative of the root’s interior is $2x$.  It was not the exterior $x^2$, but perhaps integration by parts would work.  I tried $u=x \longrightarrow u'=dx$ and $v'=x\sqrt{1+x^2} dx \longrightarrow v=\frac{1}{2} \left( 1+x^2 \right)^{3/2} \cdot \frac{2}{3}$.  Rewriting the original integral gave

The remaining integral still suggested a trig substitution, so I again tried $x =tan \theta$ to get

but the odd number of secants led me to the same dead end from trigonometric identities that stopped my original attempt.  I tried a few other variations on these themes, but nothing seemed to work.  That’s when I wondered if the integral even had a closed form solution.  Lots of simple looking integrals don’t work out nicely; perhaps this was one of them.  Plugging the integral into my Nspire CAS gave the following.

OK, now I was frustrated.  The solution wasn’t particularly pretty, but a closed form definitely existed.  The logarithm was curious, but I was heartened by the middle term I had seen with a different coefficient in my integration by parts approach.  I had other things to do, so I employed another good problem solving strategy:  I quit working on it for a while.  Sometimes you need to allow your sub-conscious to chew on an idea for a spell.  I made a note about the integral on my To Do list and walked away.

As often happens to me on more challenging problems, I woke this morning with a new idea.  I was still convinced that trig substitutions should work in some way, but my years of teaching AP Calculus and its curricular restrictions had blinded me to other possibilities.  Why not try a hyperbolic trig substitution? In many ways, hyperbolic trig is easier to manipulate than circular trig.  I knew

$\frac{d}{dt}cosh(t)=sinh(t)$ and $\frac{d}{dt}sinh(t)=cosh(t)$,

and the hyperbolic identity

$cosh^2t - sinh^2t=1 \longrightarrow cosh^2t=1+sinh^2t$.

(In case you haven’t worked with hyperbolic trig functions before, you can prove these for yourself using the definitions of hyperbolic sine and cosine:  $cosh(x)=\frac{1}{2}\left( e^x + e^{-x} \right)$ and $sinh(x)=\frac{1}{2}\left( e^x - e^{-x} \right)$.)

So, $x=sinh(A) \longrightarrow dx=cosh(A) dA$, and substitution gives

Jackpot!  I was down to an even number of (hyperbolic) trig functions, so Pythagorean identities should help me revise my latest expression into some workable form.

To accomplish this, I employed a few more hyperbolic trig identities:

1. $sinh(2A)=2sinh(A)cosh(A)$
2. $cosh(2A)=cosh^2(A)+sinh^2(A)$
3. $cosh^2(A) = \frac{1}{2}(cosh(2A)+1)$
4. $sinh^2(A) = \frac{1}{2}(cosh(2A)-1)$

(All of these can be proven using the definitions of sinh and cosh above.  I encourage you to do so if you haven’t worked much with hyperbolic trig before.  I’ve always liked the close parallels between the forms of circular and hyperbolic trig relationships and identities.)

If you want to evaluate $\int x^2 \sqrt{x^2+1} dx$ yourself, do so before reading any further.

Using equations 3 & 4, expanding, and then equation 3 again turns the integral into something that can be integrated directly.

The integral was finally solved!  I then used equations 1 & 2 to rewrite the expression back into hyperbolic functions of A only.

The integral was solved using the substitution $x=sinhA \longrightarrow A=sinh^{-1}x$ and (using $cosh^2A-sinh^2A=1$), $coshA=\sqrt{x^2+1}$.  Substituting back gave:

but that didn’t match what my CAS had given.  I could have walked away, but I had to know if I had made an error someplace or just had found a different expression for the same quantity.  I knew the inverse sinh could be replaced with a logarithm via a quadratic expression in $e^x$.

Well, that explained the presence of the logarithm in the CAS solution, but I was still worried by the cubic in my second term and the fact that my first two terms were a sum whereas the CAS’s solution’s comparable terms were a difference.  But as a former student once said, “If you take care of the math, the math will take care of you.”  These expressions had to be the same, so I needed to complete one more identity–algebraic this time.  Factoring, rewriting, and re-expanding did the trick.

What a fun problem (for me) this turned out to be.  It’s absolutely not worth the effort to do this every time when a CAS or integral table can drop the solution so much more quickly, but it’s also deeply satisfying to me to know why the form of the solution is what it is.  It’s also nice to know that I found not one, but three different forms of the solution.

Morals:  Never give up.  Trust your instincts. Never give up. Try lots of variations on your instincts. And never give up!

5 responses to “Old school integral”

1. Odd powers of secant are notoriously difficult to integrate. I don’t know if it is still taught, but I recall being asked to integrate secant cubed on an exam when I took calculus. (I was unable to solve the problem.) There is an entire Wikipedia page devoted to the integral of secant cubed, but the traditional method involves integrating by parts twice and solving for the original integral.

• Thanks, David. I’ll give that a try, too. Nice to know that there _might_ be a possibility outside of hyperbolic trigonometry.

2. Pingback: Integral follow-up | CAS Musings

3. EKM

Chris,
I would start from your first formula in bold. if you OK with integrating sec(x)dx , you can integrate any odd power of sec(x). For example, sec^3(x)dx = sec^2(x)sec(x)dx=sec(x)d(tan(x))…. then do by parts and use identity to convert tan^2(x) to sec^2(x) . You will be able to express integral of sec^3(x) as a function of the integral of sec(x). As a matter of fact you will deal with integration of 5th power of secant, but the logic is the same – express that as integral of the third power, etc.
Good Luck.
EKM

4. Interesting post, showing once again that there is no one true approach to integration. Or, maybe if you’re a computer and can perform the Risch algorithm 😉
By the way, my integral calculator (based on the Maxima CAS) returns your first result:
http://www.integral-calculator.com/#expr=x%C2%B2%2Asqrt%281%2Bx%C2%B2%29
Wolfram manages to get the result a bit shorter.