Engaging students: Deriving the double angle formulas for sine, cosine, and tangent

In my capstone class for future secondary math teachers, I ask my students to come up with ideas for engaging their students with different topics in the secondary mathematics curriculum. In other words, the point of the assignment was not to devise a full-blown lesson plan on this topic. Instead, I asked my students to think about three different ways of getting their students interested in the topic in the first place.

I plan to share some of the best of these ideas on this blog (after asking my students’ permission, of course).

This student submission comes from my former student Daniel Adkins. His topic, from Precalculus: deriving the double angle formulas for sine, cosine, and tangent.

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How does this topic extend what your students should have already learned?

A major factor that simplifies deriving the double angle formulas is recalling the trigonometric identities that help students “skip steps.” This is true especially for the Sum formulas, so a brief review of these formulas in any fashion would help students possibly derive the equations on their own in some cases. Listed below are the formulas that can lead directly to the double angle formulas.

A list of the formulas that students can benefit from recalling:

  • Sum Formulas:
    • sin(a+b) = sin(a)cos(b) + cos(a)sin(b)
    • cos(a+b) = cos(a)cos(b) – sin(a)sin(b)
    • tan(a+b) = [tan(a) +tan(b)] / [1-tan(a)tan(b)]

 

  • Pythagorean Identity:
    • Sin2 (a) + Cos2(a) = 1

 

This leads to the next topic, an activity for students to attempt the equation on their own.

 

 

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How could you as a teacher create an activity or project that involves your topic?

I’m a firm believer that the more often a student can learn something of their own accord, the better off they are. Providing the skeletal structure of the proofs for the double angle formulas of sine, cosine, and tangent might be enough to help students reach the formulas themselves. The major benefit of this is that, even though these are simple proofs, they have a lot of variance on how they may be presented to students and how “hands on” the activity can be.

I have an example worksheet demonstrating this with the first two double angle formulas attached below. This is in extremely hands on format that can be given to students with the formulas needed in the top right corner and the general position where these should be inserted. If needed the instructor could take this a step further and have the different Pythagorean Identities already listed out (I.e. Cos2(a) = 1 – Sin2(a), Sin2(a) = 1 – Cos2(a)) to emphasize that different formats could be needed. This is an extreme that wouldn’t take students any time to reach the conclusions desired. Of course a lot of this information could be dropped to increase the effort needed to reach the conclusion.

A major benefit with this also is that even though they’re simple, students will still feel extremely rewarded from succeeding on this paper on their own, and thus would be more intrinsically motivated towards learning trig identities.

 

 

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How can Technology be used to effectively engage students with this topic?

When it comes to technology in the classroom, I tend to lean more on the careful side. I know me as a person/instructor, and I know I can get carried away and make a mess of things because there was so much excitement over a new toy to play with. I also know that the technology can often detract from the actual math itself, but when it comes to trigonometry, and basically any form of geometric mathematics, it’s absolutely necessary to have a visual aid, and this is where technology excels.

The Wolfram Company has provided hundreds of widgets for this exact purpose, and below, you’ll find one attached that demonstrates that sin(2a) appears to be equal to its identity 2cos(a)sin(a). This is clearly not a rigorous proof, but it will help students visualize how these formulas interact with each other and how they may be similar. The fact that it isn’t rigorous may even convince students to try to debunk it. If you can make a student just irritated enough that they spend a few minutes trying to find a way to show you that you’re wrong, then you’ve done your job in that you’ve convinced them to try mathematics for a purpose.

After all, at the end of the day, it doesn’t matter how you begin your classroom, or how you engage your students, what matters is that they are engaged, and are willing to learn.

Wolfram does have a free cdf reader for its demonstrations on this website: http://demonstrations.wolfram.com/AVisualProofOfTheDoubleAngleFormulaForSine/

 

References

Different Ways of Computing a Limit: Index

I’m doing something that I should have done a long time ago: collecting a series of posts into one single post. The following links comprised my series on different ways of computing the limit

\displaystyle \lim_{x \to \infty} \frac{\sqrt{x^2+1}}{x}

Part 1: Algebra

Part 2: L’Hopital’s Rule

Part 3: Trigonometric substitution

Part 4: Geometry

Part 5: Geometry again

 

 

Different Ways of Solving a Contest Problem: Index

I’m doing something that I should have done a long time ago: collecting a series of posts into one single post. The following links comprised my series on different ways of solving the contest problem “If 3 \sin \theta = \cos \theta, what is \sin \theta \cos \theta?”

Part 1: Drawing the angle \theta

Part 2: A first attempt using a Pythagorean identity.

Part 3: A second attempt using a Pythagorean identity and the original hypothesis for \theta.

 

 

Different ways of computing a limit (Part 3)

One of my colleagues placed the following problem on an exam for his Calculus II course…

\displaystyle \lim_{x \to \infty} \frac{\sqrt{x^2+1}}{x}

and was impressed by the variety of correct responses that he received. I thought it would be fun to discuss some of the different ways that this limit can be computed.

Method #3. A trigonometric identity. When we see \sqrt{x^2+1} inside of an integral, one kneejerk reaction is to try the trigonometric substitution x = \tan \theta. So let’s use this here. Also, since x \to \infty, we can change the limit to be \theta \to \pi/2:

\displaystyle \lim_{x \to \infty} \frac{\sqrt{x^2+1}}{x} = \displaystyle \lim_{\theta \to \pi/2} \frac{\sqrt{\tan^2 \theta+1}}{\tan \theta}

= \displaystyle \lim_{\theta \to \pi/2} \frac{\sqrt{\sec^2 \theta}}{\tan \theta}

= \displaystyle \lim_{\theta \to \pi/2} \frac{ \sec \theta}{\tan \theta}

= \displaystyle \lim_{\theta \to \pi/2} \frac{ ~~\displaystyle \frac{1}{\cos \theta} ~~}{ ~~ \displaystyle \frac{\sin \theta}{\cos \theta} ~~ }

= \displaystyle \lim_{\theta \to \pi/2} \frac{ 1}{\sin \theta}

= 1.

Different ways of solving a contest problem (Part 3)

The following problem appeared on the American High School Mathematics Examination (now called the AMC 12) in 1988:

If 3 \sin \theta = \cos \theta, what is \sin \theta \cos \theta?

When I presented this problem to a group of students, I was pleasantly surprised by the amount of creativity shown when solving this problem.

Yesterday, I presented a solution using a Pythagorean identity, but I was unable to be certain if the final answer was a positive or negative without drawing a picture. Here’s a third solution that also use a Pythagorean trig identity but avoids this difficulty. Again, I begin by squaring both sides.

9 \sin^2 \theta = \cos^2 \theta

9 (1 - \cos^2 \theta) = \cos^2 \theta

9 - 9 \cos^2 \theta = \cos^2 \theta

9 = 10 \cos^2 \theta

\displaystyle \frac{9}{10} = \cos^2 \theta

\displaystyle \pm \frac{3}{\sqrt{10}} = \cos \theta

Yesterday, I used the Pythagorean identity again to find \sin \theta. Today, I’ll instead plug back into the original equation 3 \sin \theta = \cos \theta:

3 \sin \theta = \cos \theta

3 \sin \theta = \displaystyle \frac{3}{\sqrt{10}}

\sin \theta = \displaystyle \pm \frac{1}{\sqrt{10}}

Unlike the example yesterday, the signs of \sin \theta and \cos \theta must agree. That is, if \cos \theta = \displaystyle \frac{3}{\sqrt{10}}, then \sin \theta = \displaystyle \frac{1}{\sqrt{10}} must also be positive. On the other hand, if \cos \theta = \displaystyle -\frac{3}{\sqrt{10}}, then \sin \theta = \displaystyle -\frac{1}{\sqrt{10}} must also be negative.

If they’re both positive, then

\sin \theta \cos \theta = \displaystyle \left( \frac{1}{\sqrt{10}} \right) \left( \frac{3}{\sqrt{10}} \right) =\displaystyle \frac{3}{10},

and if they’re both negative, then

\sin \theta \cos \theta = \displaystyle \left( -\frac{1}{\sqrt{10}} \right) \left( -\frac{3}{\sqrt{10}} \right) = \displaystyle \frac{3}{10}.

Either way, the answer must be \displaystyle \frac{3}{10}.

This is definitely superior to the solution provided in yesterday’s post, as there’s absolutely no doubt that the product \sin \theta \cos \theta must be positive.

Different ways of solving a contest problem (Part 2)

The following problem appeared on the American High School Mathematics Examination (now called the AMC 12) in 1988:

If 3 \sin \theta = \cos \theta, what is \sin \theta \cos \theta?

When I presented this problem to a group of students, I was pleasantly surprised by the amount of creativity shown when solving this problem.

Yesterday, I presented a solution using triangles. Here’s a second solution that I received: begin by squaring both sides and using a Pythagorean trig identity.

9 \sin^2 \theta = \cos^2 \theta

9 (1 - \cos^2 \theta) = \cos^2 \theta

9 - 9 \cos^2 \theta = \cos^2 \theta

9 = 10 \cos^2 \theta

\displaystyle \frac{9}{10} = \cos^2 \theta

\displaystyle \pm \frac{3}{\sqrt{10}} = \cos \theta

We use the Pythagorean identity again to find \sin \theta:

\displaystyle \frac{9}{10} = \cos^2 \theta

\displaystyle \frac{9}{10} = 1 - \sin^2 \theta

\sin^2 \theta = \displaystyle \frac{1}{10}

\sin \theta = \displaystyle \pm \frac{1}{\sqrt{10}}

Therefore, we know that

\sin \theta \cos \theta = \displaystyle \left( \pm \frac{1}{\sqrt{10}} \right) \left( \pm \frac{3}{\sqrt{10}} \right) = \displaystyle \pm \displaystyle \frac{3}{10},

so the answer is either \displaystyle \frac{3}{10} or \displaystyle -\frac{3}{10}. However, this was a multiple-choice contest problem and \displaystyle -\frac{3}{10} was not listed as a possible answer, and so the answer must be \displaystyle \frac{3}{10}.

green lineFor a contest problem, the above logic makes perfect sense. However, the last step definitely plays to the fact that this was a multiple-choice problem, and the concluding step would not have been possible had \displaystyle -\frac{3}{10} been given as an option.

 

How I Impressed My Wife: Part 5b

Amazingly, the integral below has a simple solution:

Q = \displaystyle \int_0^{2\pi} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x} = \displaystyle \frac{2\pi}{|b|}.

Even more amazingly, the integral Q ultimately does not depend on the parameter a. For several hours, I tried to figure out a way to demonstrate that Q is independent of a, but I couldn’t figure out a way to do this without substantially simplifying the integral, but I’ve been unable to do so (at least so far).

So here’s what I have been able to develop to prove that Q is independent of a without directly computing the integral Q.

green lineEarlier in this series, I showed that

Q = 2 \displaystyle \int_{-\pi/2}^{\pi/2} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}

= 2 \displaystyle \int_{-\pi/2}^{\pi/2} \frac{\sec^2 x dx}{1 + 2 a \tan x + (a^2 + b^2) \tan^2 x}

= 2 \displaystyle \int_{-\infty}^{\infty} \frac{du}{1 + 2 a u + (a^2+b^2) u^2}

= \displaystyle \frac{2}{a^2+b^2} \int_{-\infty}^{\infty} \frac{dv}{v^2 + \displaystyle \frac{b^2}{(a^2+b^2)^2} }

= \displaystyle 2 \int_{-\infty}^{\infty} \frac{(a^2+ b^2) dv}{(a^2 + b^2) v^2 + b^2 }

Yesterday, I showed used the substitution w = (a^2 + b^2) v to show that Q was independent of a. Today, I’ll use a different method to establish the same result. Let

Q(a) = \displaystyle 2 \int_{-\infty}^{\infty} \frac{(a^2+b^2) dv}{(a^2+b^2)^2 v^2 + b^2 }.

Notice that I’ve written this integral as a function of the parameter a. I will demonstrate that Q'(a) = 0, so that Q(c) is a constant with respect to a. In other words, Q(a) does not depend on a.

To do this, I differentiate under the integral sign with respect to a (as opposed to x) using the Quotient Rule:

Q'(a) = \displaystyle 2 \int_{-\infty}^{\infty} \frac{ 2a \left[ (a^2+b^2)^2 v^2 + b^2\right] - 2 (a^2+b^2) \cdot (a^2+b^2) v^2 \cdot 2a }{\left[ (a^2+b^2)^2 v^2 + b^2 \right]^2} dv

Q'(a) = \displaystyle 4a \int_{-\infty}^{\infty} \frac{(a^2+b^2)^2 v^2 + b^2- 2 (a^2+b^2)^2 v^2}{\left[ (a^2+b^2)^2 v^2 + b^2 \right]^2} dv

Q'(a) = \displaystyle 4a \int_{-\infty}^{\infty} \frac{b^2-(a^2+b^2)^2 v^2}{\left[ (a^2+b^2)^2 v^2 + b^2 \right]^2} dv

I now apply the trigonometric substitution v = \displaystyle \frac{b}{a^2+b^2} \tan \theta, so that

(a^2+b^2)^2 v^2 = (a^2+b^2)^2 \displaystyle \left[ \frac{b}{a^2+b^2} \tan \theta \right]^2 = b^2 \tan^2 \theta

and

dv = \displaystyle \frac{b}{a^2+b^2} \sec^2 \theta \, d\theta

The endpoints of integration change from -\infty < v < \infty to -\pi/2 < \theta < \pi/2, and so

Q'(a) = \displaystyle 4a \int_{-\pi/2}^{\pi/2} \frac{b^2- b^2 \tan^2 \theta}{\left[ b^2 \tan^2 \theta + b^2 \right]^2} \frac{b}{a^2+b^2} \sec^2 \theta \, d\theta

= \displaystyle \frac{4ab^3}{a^2+b^2} \int_{-\pi/2}^{\pi/2} \frac{[1- \tan^2 \theta] \sec^2 \theta}{\left[ \tan^2 \theta +1 \right]^2} d\theta

= \displaystyle \frac{4ab^3}{a^2+b^2} \int_{-\pi/2}^{\pi/2} \frac{[1-\tan^2 \theta] \sec^2 \theta}{\left[ \sec^2 \theta \right]^2} d\theta

= \displaystyle \frac{4ab^3}{a^2+b^2} \int_{-\pi/2}^{\pi/2} \frac{[1-\tan^2 \theta] \sec^2 \theta}{\sec^4 \theta} d\theta

= \displaystyle \frac{4ab^3}{a^2+b^2} \int_{-\pi/2}^{\pi/2} \frac{[1- \tan^2 \theta]}{\sec^2 \theta} d\theta

= \displaystyle \frac{4ab^3}{a^2+b^2} \int_{-\pi/2}^{\pi/2} [1- \tan^2 \theta] \cos^2 \theta \, d\theta

= \displaystyle \frac{4ab^3}{a^2+b^2} \int_{-\pi/2}^{\pi/2} [\cos^2 \theta -\sin^2 \theta] d\theta

= \displaystyle \frac{4ab^3}{a^2+b^2} \int_{-\pi/2}^{\pi/2} \cos 2\theta \, d\theta

= \displaystyle \left[ \frac{2ab^3}{a^2+b^2} \sin 2\theta \right]^{\pi/2}_{-\pi/2}

= \displaystyle \frac{2ab^3}{a^2+b^2} \left[ \sin \pi - \sin (-\pi) \right]

= \displaystyle \frac{2ab^3}{a^2+b^2} \left[ 0- 0 \right]

= 0.

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I’m not completely thrilled with this demonstration that Q is independent of a, mostly because I had to do so much simplification of the integral Q to get this result. As I mentioned in yesterday’s post, I’d love to figure out a way to directly start with

Q = \displaystyle \int_0^{2\pi} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}

and demonstrate that Q is independent of a, perhaps by differentiating Q with respect to a and demonstrating that the resulting integral must be equal to 0. However, despite several hours of trying, I’ve not been able to establish this result without simplifying Q first.

How I Impressed My Wife: Part 5a

Amazingly, the integral below has a simple solution:

Q = \displaystyle \int_0^{2\pi} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x} = \displaystyle \frac{2\pi}{|b|}.

Even more amazingly, the integral Q ultimately does not depend on the parameter a. For several hours, I tried to figure out a way to demonstrate that Q is independent of a, but I couldn’t figure out a way to do this without substantially simplifying the integral, but I’ve been unable to do so (at least so far).

So here’s what I have been able to develop to prove that Q is independent of a without directly computing the integral Q.

green lineEarlier in this series, I showed that

Q = 2 \displaystyle \int_{-\pi/2}^{\pi/2} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}

= 2 \displaystyle \int_{-\pi/2}^{\pi/2} \frac{\sec^2 x dx}{1 + 2 a \tan x + (a^2 + b^2) \tan^2 x}

= 2 \displaystyle \int_{-\infty}^{\infty} \frac{du}{1 + 2 a u + (a^2+b^2) u^2}

= \displaystyle \frac{2}{a^2+b^2} \int_{-\infty}^{\infty} \frac{dv}{v^2 + \displaystyle \frac{b^2}{(a^2+b^2)^2} }

I now multiply the top and bottom of this last integral by a^2 + b^2:

Q = \displaystyle \frac{2}{(a^2+b^2)^2} \int_{-\infty}^{\infty} \frac{(a^2+ b^2) dv}{v^2 + \displaystyle \frac{b^2}{(a^2+b^2)^2} }

= \displaystyle 2 \int_{-\infty}^{\infty} \frac{(a^2+ b^2) dv}{(a^2 + b^2) v^2 + b^2 }

I now employ the substitution w = (a^2 + b^2) v, so that dw = (a^2 + b^2) v. Since a^2 + b^2 > 0, the endpoints of integration do not change, and so

Q = \displaystyle 2 \int_{-\infty}^{\infty} \frac{dw}{w^2 + b^2 }.

This final integral is independent of a.

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Since Q is independent of a, I can substitute any convenient value of a that I wish. For example, I can let a = 0 without altering the value of Q:

Q = \displaystyle \int_0^{2\pi} \frac{dx}{\cos^2 x + 2 \cdot 0 \cdot \sin x \cos x + (0^2 + b^2) \sin^2 x} = \displaystyle \int_0^{2\pi} \frac{dx}{\cos^2 x + b^2 \sin^2 x}

This provides a considerable simplification of the integral Q which also opens up additional methods of evaluation.

How I Impressed My Wife: Part 3f

Previously in this series, I showed that

Q = \displaystyle \int_0^{2\pi} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x} = \displaystyle \frac{2\pi}{|b|}.

My wife had asked me to compute this integral by hand because Mathematica 4 and Mathematica 8 gave different answers. At the time, I eventually obtained the solution by multiplying the top and bottom of the integrand by \sec^2 x and then employing the substitution u = \tan x (after using trig identities to adjust the limits of integration).
But this wasn’t the only method I tried. Indeed, I tried two or three different methods before deciding they were too messy and trying something different. So, for the rest of this series, I’d like to explore different ways that the above integral can be computed.
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So far, I have shown that

Q = \displaystyle \int_0^{2\pi} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}

= \displaystyle \int_0^{2\pi} \frac{2 \, dx}{1+\cos 2x + 2 a \sin 2x + (a^2 + b^2)(1-\cos 2x)}

= 2 \displaystyle \int_0^{2\pi} \frac{d\theta}{(1+a^2+b^2) + 2 a \sin \theta + (1 - a^2 - b^2) \cos \theta}

= 2 \displaystyle \int_{0}^{2\pi} \frac{d\theta}{S + R \cos (\theta - \alpha)}

= 2 \displaystyle \int_{0}^{2\pi} \frac{d\phi}{S + R \cos \phi}

= 2 \displaystyle \int_{-\pi}^{\pi} \frac{d\phi}{S + R \cos \phi}

where R = \sqrt{(2a)^2 + (1-a^2-b^2)^2} and S = 1 + a^2 + b^2 (and \alpha is a certain angle that is now irrelevant at this point in the calculation).

There are actually a couple of ways for computing this last integral. Today, I’ll lay the foundation for the “magic substitution”

u = \tan \displaystyle \frac{\phi}{2}

With this substitution, the above integral will become a rational function, which can then be found using standard techniques.

First, we use some trig identities to rewrite \cos 2x in terms of \tan x:

\cos 2x = 2\cos^2 x - 1

= \displaystyle \frac{ \sec^2 x (2 \cos^2 x - 1)}{\sec^2 x}

= \displaystyle \frac{ 2 - \sec^2 x)}{\sec^2 x}

= \displaystyle \frac{ 2 - [ 1 + \tan^2 x])}{1 + \tan^2 x}

= \displaystyle \frac{1- \tan^2 x}{1 + \tan^2 x}

Next, I’ll replace x by \phi/2:

\cos \phi = \displaystyle \frac{1- \tan^2 (\phi/2)}{1 + \tan^2 (\phi/2)} = \displaystyle \frac{1-u^2}{1+u^2}.

Second, for the sake of completeness (even though it isn’t necessary for this particular integral), I’ll rewrite \sin 2x in terms of \tan x:

\sin 2x = 2\sin x \cos x

= \displaystyle \frac{2\sin x \cos x \sec^2 x}{\sec^2 x}

= \displaystyle \frac{ ~ \displaystyle \frac{2 \sin x}{\cos x} ~ }{\sec^2 x}

= \displaystyle \frac{ 2 \tan x }{\sec^2 x}

= \displaystyle \frac{ 2 \tan x }{1 + \tan^2 x}

= \displaystyle \frac{2 \tan x}{1 + \tan^2 x}

Next, I’ll replace x by \phi/2:

\sin \phi = \displaystyle \frac{2 \tan^2 (\phi/2)}{1 + \tan^2 (\phi/2)} = \displaystyle \frac{2u}{1+u^2}.

Third, again for the sake of completeness,

\tan \phi = \displaystyle \frac{\sin u}{\cos u} = \displaystyle \frac{ ~ \displaystyle \frac{2u}{1+u^2} ~ }{ ~ \displaystyle \frac{1-u^2}{1+u^2} ~ } = \displaystyle \frac{2u}{1-u^2}.

Finally, I need to worry about what happens to the d\phi:

u = \tan \displaystyle \frac{\phi}{2}

du = \displaystyle \frac{1}{2} \sec^2 \displaystyle \frac{\phi}{2} \, d\phi

du = \displaystyle \frac{1}{2} \left[ 1 + \tan^2 \displaystyle \frac{\phi}{2} \right] d\phi

du = \displaystyle \frac{1}{2} (1+u^2) d\phi

\displaystyle \frac{2 du}{1+u^2} = d\phi

These four substitutions can be used to convert trigonometric integrals into some other integral. Usually, the new integrand is pretty messy, and so these substitutions should only be used sparingly, as a last resort.

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I’ll continue this different method of evaluating this integral in tomorrow’s post.

How I Impressed My Wife: Part 2f

Some husbands try to impress their wives by lifting extremely heavy objects or other extraordinary feats of physical prowess.

That will never happen in the Quintanilla household in a million years.

But she was impressed that I broke an impasse in her research and resolved a discrepancy between Mathematica 4 and Mathematica 8 by finding the following integral by hand in less than an hour:

Q = \displaystyle \int_0^{2\pi} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}

In this series, I’ll explore different ways of evaluating this integral.green lineSo far in this series, I’ve shown that

Q = 2 \displaystyle \int_{-\pi/2}^{\pi/2} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}

= 2 \displaystyle \int_{-\pi/2}^{\pi/2} \frac{\sec^2 x dx}{1 + 2 a \tan x + (a^2 + b^2) \tan^2 x}

= 2 \displaystyle \int_{-\infty}^{\infty} \frac{du}{1 + 2 a u + (a^2+b^2) u^2}

= \displaystyle \frac{2}{a^2+b^2} \int_{-\infty}^{\infty} \frac{dv}{v^2 + \displaystyle \frac{b^2}{(a^2+b^2)^2} }

= \displaystyle \frac{2}{|b|} \int_{-\infty}^{\infty} \frac{ dw }{w^2 +1}

This last integral can be evaluated using a standard trick. Let \theta = \tan^{-1} w, so that w = \tan \theta. We differentiate this last equation with respect to w:

\displaystyle \frac{dw}{dw} = \sec^2 \theta \cdot \displaystyle \frac{d\theta}{dw}

Employing a Pythagorean identity, we have

1 = (1+ \tan^2 \theta) \cdot \displaystyle \frac{d\theta}{dw}

Since w = \tan \theta, we may rewrite this as

1 = (1+ w^2) \cdot \displaystyle \frac{d\theta}{dw}

\displaystyle \frac{1}{1+w^2} = \displaystyle \frac{d\theta}{dw}

\displaystyle \frac{1}{1+w^2} = \displaystyle \frac{d}{dw} \tan^{-1} w

Integrating both sides with respect to w, we obtain the antiderivative

\displaystyle \int \frac{1}{1+w^2} = \tan^{-1} w + C

We now employ this antiderivative to evaluate Q:

Q = \displaystyle \frac{2}{|b|} \int_{-\infty}^{\infty} \frac{ dw }{w^2 +1}

= \displaystyle \frac{2}{|b|} \displaystyle \left[ \tan^{-1} w \right]^{\infty}_{-\infty}

= \displaystyle \frac{2}{|b|} \displaystyle \left[ \displaystyle \frac{\pi}{2} - \frac{-\pi}{2} \right]

= \displaystyle \frac{2\pi}{|b|}

And so, at long last, we have arrived at the solution for the integral Q. Surprisingly, the answer is independent of the parameter a.

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These last few posts illustrated the technique that I used to compute this integral for my wife in support of her recent paper in Physical Review A. However, I had more than a few false starts along the way… or, at the time, I thought they were false starts. It turns out that there are multiple ways of evaluating this integral, and I’ll explore another method of attack beginning with tomorrow’s post.