How I Impressed My Wife: Part 3d

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.
green line

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)},

where R = \sqrt{(2a)^2 + (1-a^2-b^2)^2}, S = 1 + a^2 + b^2, and \alpha is a certain angle (that will soon become irrelevant).

I now write Q as a new sum Q_3 + Q_4 by dividing the region of integration:

Q_3 = 2 \displaystyle \int_{0}^{\alpha} \frac{d\theta}{S + R \cos (\theta - \alpha)},

Q_4 = 2 \displaystyle \int_{\alpha}^{2\pi} \frac{d\theta}{S + R \cos (\theta - \alpha)}.

For Q_3, I employ the substitution u = \theta + 2\pi, so that \theta = u - 2\pi and d\theta = du. Also, the interval of integration changes from 0 \le \theta \le \alpha to 2\pi \le u \le 2\pi + \alpha, so that

Q_3 = 2 \displaystyle \int_{2\pi}^{2\pi + \alpha} \frac{du}{S + R \cos (u - 2\pi - \alpha)}

Next, I employ the trigonometric identity \cos(u - 2\pi - \alpha) = \cos (u -\alpha):

Q_3 = 2 \displaystyle \int_{2\pi}^{2\pi + \alpha} \frac{du}{S + R \cos (u - \alpha)} = 2 \displaystyle \int_{2\pi}^{2\pi + \alpha} \frac{d\theta}{S + R \cos (\theta - \alpha)},

where I have changed the dummy variable from u back to \theta.

Therefore, Q = Q_4 + Q_3 becomes

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

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

Next, I employ the substitution \phi = \theta - \alpha, so that d\phi = d\theta and the interval of integration changes from \alpha \le \theta \le 2\pi + \alpha to 0 \le \phi \le 2\pi:

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

Almost by magic, the mysterious angle \alpha has completely disappeared, making the integral that much easier to compute.

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

How I Impressed My Wife: Part 3c

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.
green line

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}.

To simplify the denominator even further, I will combine the two trigonometric terms in the denominator; this is possible because the argument of both the sine and cosine functions are the same. To this end, notice that

2 a \sin \theta + (1 - a^2 - b^2) \cos \theta = R \displaystyle \left[ \frac{2a}{R} \sin \theta + \frac{1-a^2-b^2}{R} \cos \theta \right],

where

R = \sqrt{(2a)^2 + (1-a^2-b^2)^2}

Next, let \alpha be the unique angle so that

\cos \alpha = \displaystyle \frac{1-a^2-b^2}{\sqrt{(2a)^2 + (1-a^2-b^2)^2}},

\sin \alpha = \displaystyle \frac{2a}{\sqrt{(2a)^2 + (1-a^2-b^2)^2}}.

With this substitution, we find that

2 a \sin \theta + (1 - a^2 - b^2) \cos \theta = R [\cos \theta \cos \alpha + \sin \theta \sin \alpha]

= R \cos(\theta - \alpha)

Therefore, the integral Q may be rewritten as

Q = 2 \displaystyle \int_0^{2\pi} \frac{d\theta}{S + R \cos (\theta - \alpha)},

where R = \sqrt{(2a)^2 + (1-a^2-b^2)^2} and S = 1 + a^2 + b^2.

green line

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

How I Impressed My Wife: Part 3b

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.
green line
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)}

We now employ the substitution \theta = 2x, so that d\theta = 2 \, dx. Also, the limits of integration change from 0 \le x \le 2\pi to 0 \le \theta \le 4\pi, so that

Q = \displaystyle \int_0^{4\pi} \frac{d\theta}{1+\cos \theta + 2 a \sin \theta + (a^2 + b^2)(1-\cos \theta)}

= \displaystyle \int_0^{4\pi} \frac{d\theta}{(1+a^2+b^2) + 2 a \sin \theta + (1 - a^2 - b^2) \cos \theta}

Next, I’ll divide write Q = Q_1 + Q_2 by dividing the interval of integration (not to be confused with the Q_1 and Q_2 used in the previous method), where

Q_1 = \displaystyle \int_0^{2\pi} \frac{d\theta}{(1+a^2+b^2) + 2 a \sin \theta + (1 - a^2 - b^2) \cos \theta}

Q_2 = \displaystyle \int_{2\pi}^{4\pi} \frac{d\theta}{(1+a^2+b^2) + 2 a \sin \theta + (1 - a^2 - b^2) \cos \theta}

For Q_2, I employ the substitution u = \theta - 2\pi, so that \theta = u + 2\pi and du = d\theta. Under this substitution, the interval of integration changes from $2\pi \le \theta \le 4\pi$ to $0 \le u \le 2\pi$, and so

Q_2 = \displaystyle \int_{0}^{2\pi} \frac{du}{(1+a^2+b^2) + 2 a \sin (u+2\pi) + (1 - a^2 - b^2) \cos (u+2\pi)}

Next, I use the periodic property for both sine and cosine — \sin(u + 2\pi) = \sin u and \cos(u+ 2\pi) = \cos u — to rewrite Q_2 as

Q_2 = \displaystyle \int_{0}^{2\pi} \frac{du}{(1+a^2+b^2) + 2 a \sin u + (1 - a^2 - b^2) \cos u}

Except for the dummy variable u, instead of \theta, we see that Q_2 is identical to Q_1. Therefore,

Q = Q_1 + Q_2 = 2 Q_1 = 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}.

green line

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

How I Impressed My Wife: Part 3a

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.
green line
For this next technique, I begin by using the trigonometric identities

\sin^2 x = \displaystyle \frac{1-\cos 2x}{2},

\cos^2x = \displaystyle \frac{1+\cos 2x}{2},

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

Using these identities, we obtain

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

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

In this way, the exponents have been removed from the denominator, thus making the integrand somewhat less complicated.

green line

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.

How I Impressed My Wife: Part 2e

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} }

I now employ the substitution v = \displaystyle \frac{b}{a^2+b^2} w, so that dv = \displaystyle \frac{b \, dw}{a^2 + b^2}. If b > 0, then the interval of integration does not change under this substitution, and so

Q = \displaystyle \frac{2}{a^2+b^2} \int_{-\infty}^{\infty} \frac{ \displaystyle \frac{b \, dw}{a^2 + b^2}}{\displaystyle \left( \displaystyle \frac{b}{a^2+b^2} w \right)^2 + \displaystyle \frac{b^2}{(a^2+b^2)^2} }

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

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

On the other hand, if b < 0, then the interval of integration does not change under this substitution but the endpoints get flipped:

Q = \displaystyle \frac{2}{a^2+b^2} \int_{\infty}^{-\infty} \frac{ \displaystyle \frac{b \, dw}{a^2 + b^2}}{\displaystyle \left( \displaystyle \frac{b}{a^2+b^2} w \right)^2 + \displaystyle \frac{b^2}{(a^2+b^2)^2} }

Q = \displaystyle -\frac{2}{a^2+b^2} \int_{-\infty}^{\infty} \frac{ \displaystyle \frac{b \, dw}{a^2 + b^2}}{\displaystyle \left( \displaystyle \frac{b}{a^2+b^2} w \right)^2 + \displaystyle \frac{b^2}{(a^2+b^2)^2} }

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

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

I can consolidate these two cases by writing

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

Clearly, the integral diverges if b = 0, and so I’ll ignore this special case from now on.

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I’m almost done; I’ll complete the evaluation of this integral in tomorrow’s post.

How I Impressed My Wife: Part 2d

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}

To evaluate this last integral, I complete the square in the denominator. I first factor (a^2+b^2) out of the denominator:

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

$latex  = \displaystyle \frac{2}{a^2+b^2} \int_{-\infty}^{\infty} \frac{du}{\left( u^2 + \displaystyle \frac{2 a}{a^2+b^2} u + \displaystyle \frac{a^2}{(a^2+b^2)^2} \right) + \displaystyle \frac{1}{a^2+b^2} – \displaystyle \frac{a^2}{(a^2+b^2)^2} }$

$latex  = \displaystyle \frac{2}{a^2+b^2} \int_{-\infty}^{\infty} \frac{du}{\left( u + \displaystyle \frac{a}{a^2+b^2} \right)^2 + \displaystyle \frac{a^2+b^2}{(a^2+b^2)^2} – \displaystyle \frac{a^2}{(a^2+b^2)^2} }$

$latex  = \displaystyle \frac{2}{a^2+b^2} \int_{-\infty}^{\infty} \frac{du}{\left( u + \displaystyle \frac{a}{a^2+b^2} \right)^2 + \displaystyle \frac{b^2}{(a^2+b^2)^2} }$

Next, I employ the substitution v = u + \displaystyle \frac{a}{a^2+b^2}, so that dv = du. The endpoint of the integral do not change with this substitution, and so

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

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I’ll continue with the evaluation of this integral in tomorrow’s post.

How I Impressed My Wife: Part 2c

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}

I now employ the substitution u = \tan x, so that du = \sec^2 x dx. Also, the endpoints change from -\pi/2 < x < \pi/2 to -\infty < u < \infty, so that

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

green line

I’ll continue with the evaluation of this integral in tomorrow’s post.

How I Impressed My Wife: Part 2b

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 lineIn yesterday’s post, 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}

We now multiply the top and bottom of the integrand by \sec^2 x. This is permissible because \sec^2 x is defined on the interior of the interval (-\pi/2, \pi/2) — which is why I needed to adjust the limits of integration in the first place. I obtain

Q = 2 \displaystyle \int_{-\pi/2}^{\pi/2} \frac{\sec^2 x dx}{\cos^2 x \sec^2 x + 2 a \sin x \cos x \sec^2 x + (a^2 + b^2) \sin^2 x \sec^2 x}
Next, I use some trigonometric identities to simplify the denominator:
  • \cos^2 x \sec^2 x = \cos^2 x \displaystyle \frac{1}{\cos^2 x} = 1
  • \sin x \cos x \sec^2 x = \sin x \cos x \frac{1}{\cos^2 x} = \displaystyle \frac{\sin x}{\cos x} = \tan x
  • \sin^2 x \sec^2 x = \sin^2 x \displaystyle \frac{1}{\cos^2 x} = \displaystyle \left( \frac{\sin x}{\cos x} \right)^2 = \tan^2 x

Therefore, the integral becomes

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

green line

I’ll continue with the evaluation of this integral in tomorrow’s post.

How I Impressed My Wife: Part 2a

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 lineI begin by adjusting the range of integration:

Q = Q_1 + Q_2 + Q_3,

where

Q_1 = \displaystyle \int_0^{\pi/2} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x},

Q_2 = \displaystyle \int_{\pi/2}^{3\pi/2} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x},

Q_3 = \displaystyle \int_{3\pi/2}^{2\pi} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}.

I’ll begin with Q_3 and apply the substitution u = x - 2\pi, or x = u + 2\pi. Then du = dx, and the endpoints change from 3\pi/2 \le x 2\pi to -\pi/2 \le u \le 0. Therefore,

Q_3 = \displaystyle \int_{-\pi/2}^{0} \frac{du}{\cos^2 (u+2\pi) + 2 a \sin (u+2\pi) \cos (u+2\pi) + (a^2 + b^2) \sin^2 (u+2\pi)}.

Next, we use the periodic property for both sine and cosine — \sin(x + 2\pi) = \sin x and \cos(x + 2\pi) = \cos x — to rewrite Q_3 as

Q_3 = \displaystyle \int_{-\pi/2}^{0} \frac{du}{\cos^2 u + 2 a \sin u \cos u + (a^2 + b^2) \sin^2 u}.

Changing the dummy variable from u back to x, we have

Q_3 = \displaystyle \int_{-\pi/2}^{0} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}.

Therefore, we can combined Q_3 + Q_1 into a single integral:

Q_3 + Q_1 = \displaystyle \int_{-\pi/2}^{0} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}

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

Q_3 + Q_1 = \displaystyle \int_{-\pi/2}^{\pi/2} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}

Next, we work on the middle integral Q_2. We use the substitution u = x - \pi, or x = u + \pi, so that du = dx. Then the interval of integration changes from \pi/2 \le x \le 3\pi/2 to -\pi/2 \le u \le \pi/2, so that

Q_2 = \displaystyle \int_{\pi/2}^{3\pi/2} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}

= \displaystyle \int_{-\pi/2}^{\pi/2} \frac{du}{\cos^2 (u+\pi) + 2 a \sin (u+\pi) \cos (u+\pi) + (a^2 + b^2) \sin^2 (u+\pi)}.

Next, we use the trigonometric identities

\sin(u + \pi) = \sin u \cos \pi + \cos u \sin \pi = \sin u \cdot (-1) + \cos u \cdot 0 = - \sin u,

\cos(u + \pi) = \cos u \cos \pi - \sin u \sin \pi = \cos u \cdot (-1) - \sin u \cdot 0 = - \cos u,

so that the last integral becomes

Q_2 = \displaystyle \int_{-\pi/2}^{\pi/2} \frac{dx}{(-\cos u)^2 + 2 a (-\sin u)(- \cos u) + (a^2 + b^2) (-\sin u)^2}

= \displaystyle \int_{-\pi/2}^{\pi/2} \frac{du}{\cos^2 u + 2 a \sin u \cos u + (a^2 + b^2) \sin^2 u}

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

On the line above, I again replaced the dummy variable of integration from u to x. We see that Q_2 = Q_1 + Q_3, and so

Q = Q_1 + Q_2 + Q_3

Q = 2 Q_2

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}

green line

I’ll continue with the evaluation of this integral in tomorrow’s post.

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