Take the 2-minute tour ×
Mathematics Educators Stack Exchange is a question and answer site for those involved in the field of teaching mathematics. It's 100% free, no registration required.

Motivating what is often called "Calculus 2" can be hard, which is probably why there are multiple other attempts at motivating it here. I have just begun teaching such a course, beginning with the bag-of-tricks for performing integration.

We will be spending a large amount of time on trig substitution, by which I mean evaluating integrals like

$$ \int \frac{1}{\sqrt{a^2 - x^2}} \mathrm d x$$

by substituting $x = a \sin \theta$, and similar integrals of that type. This was similarly emphasized when I first learned calculus years ago, but though I can defend many aspects of a "calculus 2" course in terms of use or intellectual interest, I don't have a repository of interesting or motivating reasons why we spend so much time on this subject.

In principle, these types of problems might come out of physical phenomena, as these are very Pythagoreanesque.

So I wonder:

What are specific applications and motivations for learning trig substitution?

I have started with some simple applications - but I'm really hoping for more.

share|improve this question
    
"I don't have a repository of motivating reasons why we spend so much time on this subject" -- agreed. I look forward to the day when that time is much reduced. –  Matt F. Sep 4 at 2:16
    
Apparently, some of the historical interest in these integrals is tied to the development of classical algebraic geometry. The $t=\tan(x/2)$ substitution has application to find the rational points on a circle. See pages 5 and 6 of amazon.com/Algebraic-Geometry-An-Introduction-Universitext/dp/…. Beyond interesting history, all you need is a quadratic in a model and a derivative and there it is, to solve it you integral a root. Beyond this, these are far more interesting when taught in parallel with hyperbolic substitutions. See section 9.3 of... –  James S. Cook Sep 4 at 3:32
    
of supermath.info/OldschoolCalculusII.pdf where I compare and contrast some integrals from the trigonometric verses hyperbolic technique. Btw, variational calculus is full of differential equations whose explicit solution rest on the solution of integrals with roots. –  James S. Cook Sep 4 at 3:34

5 Answers 5

In my opinion, trig substitution is presented in a terrible fashion in every calculus book I have ever seen. "If you see $\sqrt{a^2 - x^2}$, substitute $x = a \sin \theta$, and then use such-and-such trig identity, blah, blah, blah..." Yet another unmotivated rule to memorize.

I always present trig substitution as follows: If you see any algebraic expression that looks like the Pythagorean theorem (i.e., $\sqrt{a^2 - x^2}$, $\sqrt{x^2 - a^2}$, or $\sqrt{x^2 + a^2}$), then DRAW A TRIANGLE such that the algebraic expression is one of the side lengths, choose an angle $\theta$, use the triangle to write out the values of trigonometric functions involving $\theta$, and substitute.

This takes three unmotivated "rules" off the table, and gives the students a different rule in their place: when in doubt, appeal to geometry.

share|improve this answer
    
And I used to add: "Never, never, try to memorize the 'correct' substitution". –  Bernard Masse Sep 4 at 20:06

There are some areas that are naturally calculated with trig substitution, and which appear somewhat naturally. For example,

  1. Finding the area of the ellipse $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ can be done with trig substitution.
  2. Given two intersecting circles in the plane, finding the area enclosed within one circle but outside the other can be done with trig substitution. This shape is a lune.
  3. Finding the area within a circle and under a specified height leads to a trig sub integral.
  4. Equivalently, finding the amount of soda in a cylindrical can lying on its side given the height of the soda leads to a trig sub.
  5. A torus can be constructed by rotating a circle around the origin as a volume of revolution. Finding the area of this torus leads to a trig sub.

In general, finding areas of regions bounded by quadratic algebraic curves might often lead to integrals that can be tackled with trig substitution.

share|improve this answer

Although it's not particularly specific, given a rational function $F(\sin(x), \cos(x))$, the substitution $z = \tan\frac{x}{2}$ will transform the integral $\displaystyle \int F(\sin x, \cos x) \mathrm d x$ into an integral of a rational function of $z$, which can in turn (and in principle) always be integrated with partial fractions.

These come up when calculating the average angular velocity of the output shaft of a universal joint when the two shafts are not aligned (as in the picture on wikipedia).

Unfortunately, I do not have intuition for coming up with interesting questions yielding rational functions of $\sin x$ and $\cos x$ (even though I have actually used this before!), so I cannot list more.

share|improve this answer

One possible motivation originates in a very simple problem. Suppose a boat located at the origin is tied to a buoy located at $(0, l)$ (and where the rope starts out taut). If the boat travels upwards on the $y$-axis, the buoy will travel in a path described by a function $f(x)$.

Since the buoy is always pulled in the direction of the boat, the line from the boat to the buoy is a tangent to $f(x)$, and so we can write down the differential equation

$$ f'(x) = \frac{-\sqrt{l^2 - x^2}}{x},$$

which is clearly separable and clearly leads to a classical trig substitution problem. $\diamondsuit$

Similarly, any object pulling other free-moving objects with rigid materials of fixed length lead to isomorphic questions.

share|improve this answer

Although beyond the level of most students the first time they're learning trig substitution, a variety of problems asking about the electric field around uniformly charged 1-dimensional lines in space end up yielding integrals that should be done with trig substitution.

For instance, a uniformly charged rod of length $L$ lying with one end at the origin and the other at the point $(0,l)$ produces an electric field at the point $(x,y)$ given by

$$\int_{-x}^{l-x} \frac{y}{4\pi \epsilon_0(t^2 + y^2)^{3/2}} \mathrm{d}t.$$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.