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Tricks of Integration (1) – The Koop Subs

四月 8, 2009

In this article, I will talk about some of my favorite substitutions for integration, which I will named them as the “Koop subs”. They are particularly useful in mathematics competitions such as the Putnam competition.

Example 1) Compute the integral: I = \int_{0}^{1} \frac{x^3}{1 - 3x + 3x^2} dx.

The usual way to do this come in two steps:
(i) Do long division to the rational function.
(ii) Complete the square for the denominator and use a trig substitution.

Here is a smarter way:
Note that 1 - 3x + 3x^2 = x^3 + (1 - x)^3.
Making the substitution u = 1 - x yields I = \int_{0}^{1} \frac{(1 - u)^3}{u^3 + (1 - u)^3}dx.
Since it is a definite integral, u is a dummy variable and we may thus rename the variable to be x.
Hence I = \int_{0}^{1} \frac{(1 - x)^3}{x^3 + (1 - x)^3}dx.
Adding the original I to this I yields:

2I = \int_{0}^{1} \frac{x^3}{x^3 + (1 - x)^3}dx + \int_{0}^{1} \frac{(1 - x)^3}{x^3 + (1 - x)^3}dx = \int_{0}^{1} 1 dx = 1.
(Note how the numerator and denominator magically cancel each other after adding the two integrals.)
Therefore, I = \frac{1}{2}.

Example 2) Compute I = \int_{0}^{\pi/2} \sin^2{x} dx.

The usual way to compute I is by trig identity.

Here is another way, let u = \pi/2 - x yields: I = \int_{0}^{\pi/2} \cos^2{u} du. Since \sin(\pi/2 - x) = \cos(x), and \cos(\pi/2 - x) = \sin(x). Again, u is a dummy variable, therefore, we have I = \int_{0}^{\pi/2} \cos^2{x}dx.

Adding the two integrals gives: 2I = \int_{0}^{\pi/2} \sin^2{x} + \cos^2{x} dx = \int_{0}^{\pi/2} 1 dx = \pi/2.
Therefore, I = \pi/4.

Example 3) Compute I = \int_{0}^{\pi/2} \frac{1}{1 + \tan^3{x}}dx.

One way to do this is by the Weierstrass substitution, t = \tan(x/2), and then evaluate the rational function by partial fractions …etc.

Here is another way, the substitution x \mapsto \pi/2 - x gives I = \int_{0}^{\pi/2} \frac{\tan^{3}{x}}{1 + \tan^3{x}}dx.
Adding the two integrals give 2I = \int_{0}^{\pi/2} 1 dx = \pi/2. Therefore, I = \pi/4.

Notice that the number 3 in the problem is not at all special and the same technique will yield that:

I = \int_{0}^{\pi/2} \frac{1}{1 + (\tan{x})^{y}}dx = \frac{\pi}{4}. for all y \in \mathbb{R}.

In particular, if y = \sqrt{2}, then the indefinite integral cannot be represented as functions we know how to write down, which makes our method particularly effective in such cases.

Example 4) Compute the integral I = \int_{0}^{\infty} \frac{\log{x}}{1 + x^2}dx.

We let u = 1/x, then we have du/dx = -1/x^2 = -u^2. Hence I = \int_{0}^{\infty} \frac{\log{1/u}}{1 + (1/u)^2} \frac{1}{u^2}du = -\int_{0}^{\infty} \frac{\log{u}}{1 + u^2}dx = -I. Therefore, we have 2I = 0 which implies I = 0.

In general, the interval of integration often suggests what substitution to use. The idea here is to make a substitution such that the interval of integration is invariant.

For example, if we integrate from 0 to 1, then the koop sub u = 1 - x will make the interval invariant. If we integrate from 0 to \infty or from 1 to \infty, the koop sub u = 1/x will make the interval invariant. Finally, if we integrate from -1 to 1, then the koop sub u = -x will make the interval invariant.

To summarize, the types of koop sub usually used is:
1) u = 1 - x;
2) u = \pi/2 - x;
3) u = 1/x;
4) u = -x;
5) u = x - 1/x,
or some variant of the above.

Exercise: Let a > 0, compute the integral I = \int_{0}^{\infty} \frac{\log{x}}{a^2 + x^2}dx.

Coming up: Tricks of Integration (2) – Feynman’s Favorite Trick

Posted in Calculus, Teaching |

9 comments

  1. Could you explain what you mean by “invariant”, and why that is what we want to do?

         by DJ 四月 12, 2009 at 10:24 pm
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  2. Basically, all of these problems have a common theme: There is a hidden “symmetry”.

    The substitution I made helps to reveal the symmetry.

    By invariant, I mean the interval of integration is unchanged. For example, suppose we integrate from 0 to 1. If we let substitution u = 1 – x, when x = 0, 1, u = 1, 0 respectively. We then have to integrate from 1 to 0. However, du = -dx, and integrating form 1 to 0 is the same as the negative of the integral from 0 to 1. Therefore, the interval of integration is unchanged.

    Does this make sense?

         by koopakoo 四月 13, 2009 at 4:08 am
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  3. In the first trick, you start by observing that 1-3x+3x^2 = x^3+(1-x)^3. It occurs to me now that I would never have recognized that. I’ve never looked at a quadratic and said “hey, this looks like the following cubic…”. Is there some trick to making that observation? Do you have any practice problems (other than homework) we can work on using these tricks? That would really help.

         by DJ 四月 15, 2009 at 7:33 am
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  4. I call that “ingenuity” =) You will get used to it. Keep an open mind and always look for better ways to solve problems. All of a sudden, in some day, you will reach that level.

         by koopa 四月 15, 2009 at 1:00 pm
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  5. Besides, if you were to make the substitution, u = 1 – x, and try to compute 1 – 3x + 3x^2 = 1 – 3(1 – u) + 3(1 – u)^2.
    You will get that it’s equal to: 1 – 3u + 3u^2, which is the same, and you should be able to work out the same thing. The observation that it’s equal to (1 – x)^3 + x^3 only makes the problem more apparent.

         by koopakoo 四月 15, 2009 at 1:06 pm
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  6. Anyways, I will give you more problems to practice, see you at the review session.

         by koopakoo 四月 15, 2009 at 1:07 pm
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  7. What will be the Chinese name of the Koop Subs?

         by Vincent Tam 八月 12, 2009 at 3:43 pm
    Reply

    • 直譯是「古氏代換法」?!

           by myisland8132 八月 16, 2009 at 3:21 am
      Reply

  8. Not a bad name, but perhaps you can come up with a better name.

         by koopakoo 八月 16, 2009 at 12:28 pm
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