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Studying for an Integration Bee

Hello friends! My university's math club is holding an integration bee in March, and I'm excited to participate in my first one!

I was sifting through their Facebook page for posts about last year's integration bee. The integration techniques they said all participants should know were:

  • Polynomial / trig / inverse trig / exponential / rational / radical functions
  • U-substitution
  • Integration by parts
  • Partial fractions
  • Powers of trig functions
  • Trig substitution
  • Note: All integrals are single-variable.

They also take pics of a few integrals from the bee:

\[\displaystyle \int_0^{\infty} \frac{dx}{\left(x + \frac 1x\right)^2}\]

and the winning integral (which I believe is adapted from an integral from MIT's 2012 bee):

\[\int_1^2 \frac{\ln x}{2 - 2x + x^2} \ dx\]

As a restraint, I scored a 5 on my AP Calculus BC exam last year. And so I ask: how can I study for an integration bee? And not only that, but what are some essential tricks/shortcuts/strategies/generalizations I can learn?

So far in my studying, I've derived three generalizations:

\[\int x^ne^x \ dx = n!e^x \left( \frac{x^n}{n!} - \frac{x^{n - 1}}{(n - 1)!} + \frac{x^{n - 2}}{(n - 2)!} - \frac{x^{n - 3}}{(n - 3)!} + \cdots + 1\right) + C\]

\[\begin{align} \int \frac{1}{(x + a)(x + b)} \ dx = \frac{1}{a - b} \ln \left| \frac{x + b}{x + a} \right| + C, & & a > b \end{align}\]

\[\int \frac{1}{x^a + x} \ dx = -\frac{1}{a - 1}\ln \left(1 + \frac{1}{x^{a-1}}\right) + C\]

Thanks, everyone! Any and every piece of advice is appreciated! :)

Note by Zach Abueg
1 month ago

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As a start: Look up Integration tricks, Laplace transform, contour integration, hyperbolic trigonometric substitution, half angle tangent substitution, differentiation under the integral sign.

If you want some nefarious questions, go to the search bar, type "Brilliant integration contest", under the "Notes" tab.

Pi Han Goh - 1 month ago

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Thanks, Pi! Differentiation under the integral and half-angle tangent sub do wonders. I'm worried that Laplace transform, contour integration, or hyperbolic trigonometric substitution are out of my scope right now, but I'll do my best!

Zach Abueg - 1 month ago

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Remember the change of variable trick. It can come up and be very sneaky. Here's a classic example:

\[\displaystyle \int_0^{\frac{\pi}{2}} \dfrac{1}{1+\tan^{2017} (x)} \; \mathrm{d}x\]

I won't tell you how to solve, but I can give you a hint if your require it.

Sharky Kesa - 1 month ago

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Seeing a large power like that, my first instinct is to generalize, so we must have

\[\begin{align} I_n & = \int_0^{\frac{\pi}{2}} \frac{1}{1 + \tan ^n (x)} \ dx & \small \color{blue} \text{Multiplying through by } \cos ^n x \\ & = \int_0^{\frac{\pi}{2}} \frac{\cos ^n x}{\cos ^n x + \sin ^n x} \ dx \end{align}\]


Edit: I got it! Those limits are so sneaky! We use \(\displaystyle \int_a^b f(x) \ dx = \int_a^b f(a + b - x)\). Knowing that \(\cos \left(\frac{\pi}{2} - x\right) = \sin x\) and vice versa, we see that

\[\begin{align} I_n & = \int_0^{\frac{\pi}{2}} \frac{\cos ^n x}{\cos ^n x + \sin ^n x} \ dx \\ & = \int_0^{\frac{\pi}{2}} \frac{\cos ^n \left(\frac{\pi}{2} - x\right)}{\cos ^n \left(\frac{\pi}{2} - x\right) + \sin^n \left(\frac{\pi}{2} - x \right)} \ dx \\ & = \int_0^{\frac{\pi}{2}} \frac{\sin ^n x}{\sin ^n x + \cos ^n x} \ dx \\ \implies 2I_n & = \frac{\pi}{2} \\ \implies I_n & = \frac{\pi}{4} & \small \color{blue} \text{A result independent of } n \end{align}\]

Oh my! That's amazing!

Zach Abueg - 1 month ago

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Hey thanks for tagging me!

I've never actually done an integration contest, but I am pretty familiar with the math contest style integrals where the solution is usually rather straightforward if you look at the problem in a certain way.

SPOILER ALERT! Solution incoming

For example, the first integral you posted I was able to do by using some symmetry and then a trig substitution. It might not be the easiest way, but the steps weren't some crazy substitution you pull out of nowhere. They 'make sense'

First, I pulled out an \(x^{2}\) from the denominator;

\[ I = \int_{0}^{\infty} \frac{dx}{x^{2}(1+\frac{1}{x^{2}})^{2}}\]

this leads very nicely into one of the methods mentioned in the Integration Tricks wiki Pi Han Goh mentioned. We let \(y=\frac{1}{x}\), and so \(dy = -\frac{1}{x^{2}}dx\). This substitution flips the bounds of integration, and gets us;

\[ I = \int_{\infty}^{0} \frac{-dy}{(1+y^{2})^{2}} = \int_{0}^{\infty} \frac{dy}{(1+y^{2})^{2}} \]

My first time going through the problem, I then made the trig sub \(x= \tan(u)\), since the trig identity \(1+\tan^{2}(\theta)=\sec^{2}(\theta)\) and the fact that \(dx = \sec^{2}(\theta)d \theta\) makes the entire expression condense very nicely into a simple trig integral.

However, on my second time looking at the problem, I noticed that our form for \(I\) in terms of \(y\) looks extremely similar to the original form of the integral, so I revisited that expression again.

\[ I = \int_{0}^{\infty} \frac{dx}{(x+\frac{1}{x})^{2}} = \int_{0}^{\infty} \frac{dx}{\frac{1}{x^{2}}(1+x^{2})^{2}} = \int_{0}^{\infty} \frac{x^{2}dx}{(1+x^{2})^{2}} \]

We now add those two integrals together;

\[ 2I = \int_{0}^{\infty} \frac{x^{2}+1}{(x^{2}+1)^{2}} dx = \int_{0}^{\infty} \frac{dx}{x^{2}+1}\]

You may be already familiar with this integral, but in case not, we use the substitution \(x = \tan(u)\) to get that \(dx = \sec^{2}(u) du\) and that \(1+x^{2} = \sec^{2}(u)\). This means that our integral turns into;

\[ 2I = \int_{u_{1}}^{u_{2}} \frac{\sec^{2}(u)}{\sec^{2}(u)} du = \int_{u_{1}}^{u_{2}} du = u_{2}-u_{1} \]

Substituting back in for \(x\), the improper integral becomes;

\[ 2I = \tan^{-1}(\infty)-\tan^{-1}(0) = \frac{\pi}{2} \Rightarrow I = \frac{\pi}{4} \]


A long while back I actually posted a problem using a very similar symmetry trick, so here's a problem to replace the one I took.

I guess a general tip from me would be to go over the integration tricks wiki and understand exactly why we do these things, in particular what about the form of the integral allows these tricks to be useful. Its similar to having a toolbox, and knowing exactly what job each tool is supposed to perform.

Another helpful thing is to write your own problems using these methods. Your goal is to get really hard contest problems right, and so getting inside the head of problem writers is a great way to understand what they're trying to get you to do. I particularly like writing different versions of problems I struggle with, since this backwards thinking makes you ask yourself 'what needs to happen for this trick to be viable?'

I'm not at the IMO level, so take this with a grain of salt, but these definitely helped me out significantly in High School math contests as well as just my math/physics classes in general. We'll see how well these work in the UI Undergrad contest tomorrow ;)

Brandon Monsen - 1 month ago

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Hey Brandon, thanks for your advice! And what a great realization, indeed! Reminds me of the classic IBP integral \(\displaystyle \int e^x \cos x \ dx\) and its reappearance from solving. I love that last part of your advice and how you advise really learning how to solve integrals by creating them myself. Similar to learning a concept by teaching it! Thank you so much! By the way, how'd your contest go?

Zach Abueg - 1 month ago

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Try this one:

\[ I = \int_{0}^{e} \sqrt{1-\ln(x)} \ dx\]

Brandon Monsen - 1 month ago

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Will do! I have some homework to catch up on, but definitely in my free time!

Zach Abueg - 1 month ago

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Hey, Brandon! Back at it, and I fumbled with this one for a while before solving it with the erf function. I saw that you posted it as a problem and your solution is great! Unfortunately, I am not at all familiar with the gamma function and its integral representations.

Zach Abueg - 1 month ago

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@Zach Abueg Gaussian integral* I mean.

Zach Abueg - 4 weeks, 1 day ago

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@Zach Abueg Gamma function is something I can definitely stand to learn much better as well. I've never heard of the erf function before though. How did you end up solving it?

Brandon Monsen - 4 weeks, 1 day ago

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@Brandon Monsen The error function, \(\text{erf } (x)\), is just a special integral, like the Gaussian integral is, to my understanding. It is defined as

\[\text{erf}(x) = \frac{2}{\sqrt{\pi}} \int_0^x e^{-t^2} \ dt\]

I made the substitution \(u = 1 - \ln x \implies x = e^{1 - u}\), solved the subbed integral by parts, and then turned the resulting integral from IBP into a form of the error function that made it easier to evaluate.

Zach Abueg - 4 weeks, 1 day ago

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@Zach Abueg ah and then I'm guessing that you used the fact that \(\text{erf}(\infty)=1\) when you plugged in the bounds?

Brandon Monsen - 4 weeks, 1 day ago

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@Brandon Monsen Right on.

Zach Abueg - 4 weeks, 1 day ago

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@Zach Abueg Ok I see that now. As for the contest, it was a lot harder than I thought. I haven't ever done a proof based contest before, and so having to write up a formal proof for every problem was a huge challenge for me, since my solutions on Brilliant tend to be more casual/informal. It also didn't help that there was quite a bit of number theory and probability on there :P (those are my worst two math subjects) Ended up at the average score. Hoping to bring that up for Virginia Tech Regionals and up a lot for the Putnam in December

Brandon Monsen - 4 weeks, 1 day ago

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@Brandon Monsen Ah, I see. That's certainly true about the proofs here. Occasionally we see the truly rigorous proof. I do a Logic and Proof course this spring, so we'll see how that goes. But hey, best of luck man! Counting on ya to kill it :)

Zach Abueg - 3 weeks, 5 days ago

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Anyone isn't mentioned. Try to edit your comment. :)

Munem Sahariar - 1 month ago

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