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! :)

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

## Comments

Sort by:

TopNewestAs 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.

Log in to reply

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!Log in to reply

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.

Log in to reply

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

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

Hm.

Edit: I got it! Those limits are

sosneaky! 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!

Log in to reply

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 incomingFor 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;

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

END SOLUTION(s)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

whywe 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 ;)

Log in to reply

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?

Log in to reply

Try this one:

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

Log in to reply

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

Log in to reply

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.

Log in to reply

Log in to reply

Log in to reply

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

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.

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

If I may, I would love to ask you guys, the very brightest of Brilliant, for your generous advice :)

@Calvin Lin

@Eli Ross @Mark Hennings @Rishabh Cool @Chew-Seong Cheong @Brian Charlesworth @Pi Han Goh @敬全 钟 @Aditya Narayan Sharma @Jon Haussmann @Jake Lai @Efren Medallo @Arjen Vreugdenhil @Brandon Monsen @Otto Bretscher @Alan Enrique Ontiveros Salazar @Sharky Kesa @Hummus a @Tapas Mazumdar @Kushal Bose @Rahil Sehgal

@Guillermo Templado @Guilherme Niedu @Hosam Hajjir

Log in to reply

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

Log in to reply