De Moivre's Theorem: Explaining Integrals of sin^mxcos^nx

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Homework Help Overview

The discussion revolves around De Moivre's Theorem and its application to evaluating integrals of the form \(\int \sin^m x \cos^n x \, dx\). Participants explore the theorem's implications in the context of trigonometric integrals and the potential methods for simplification.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Some participants attempt to explain De Moivre's Theorem and its proof, while others question its relevance to the integral evaluation. There are suggestions to use identities and the binomial theorem for expanding trigonometric functions. Additionally, participants discuss the implications of the theorem in terms of complex numbers and their geometric interpretations.

Discussion Status

The discussion is active, with various interpretations and approaches being explored. Some participants express uncertainty about the theorem's utility in solving the integral, while others suggest potential methods for simplification. There is no explicit consensus, but several productive lines of reasoning are being examined.

Contextual Notes

Participants note the complexity of the integrals involved and the challenges posed by different cases of \(m\) and \(n\). There is also mention of the need for a deeper understanding of the theorem's implications in calculus.

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Can someone explain the De Moivre's Theorem and how its used to evaluate integrals of the type \int sin^mxcos^nxdx
 
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De Moivre's Theorem:

(\cos x + i \sin x)^n = \cos nx + i \sin nx

Proof by induction or Euler's Formula.

Explain it? There it is, no intuitive feel or understanding about it, not unless you're Gauss.

How its used to evaluate integrals like that? I don't think it is..

If m is odd, let w = cos x. If n is odd, let w = sin x. If both m and n are even and non-negative, convert all to sin x or all to cos x using Pythagorean Identities and then use the 1st or 2nd integrals I list below. If m and n are even and one of them is negative, convert to which function is in the denominator and use 3 and 4 i list below. The case in which both m and n are negative and even is too hard for my brain.

1. \int \sin^n x dx = -\frac{1}{n} \sin^{n-1} x \cos x + \frac{n-1}{n}\int \sin^{n-2} x dx, n positive.

2. \int \cos^n x dx = \frac{1}{n} \sin^{n-1} x \sin x + \frac{n-1}{n}\int \cos^{n-2} x dx, n positive.

3. \int \frac{1}{\sin^m x} dx = \frac{-1}{m-1} \frac{\cos x}{\sin^{m-1} x} + \frac{m-2}{m-1}\int \frac{1}{\sin^{m-2} x} dx, m can not equal 1, positive.

4. \int \frac{1}{\cos^m x} dx = \frac{1}{m-1} \frac{\sin x}{\cos^{m-1} x} + \frac{m-2}{m-1}\int \frac{1}{\cos^{m-2} x} dx, m can not equal 1, positive.Note you may need to use those listed integrals several times for some applications. These are very long and arduous formulas to memorize or to even use. Better to get good at integration.
 
NO! FORGET EVERYTHING I SAID IN THE LAST POST!

You can use De Moivre's theorem to work out expressions for cos^n x and sin^n x into lower degrees, and work them easily from there.
 
I don't really know how De Moivre's Theorem would help... but my guess is, you use:
\sin^n(x)=\left (\frac{e^{ix}+e^{-ix}}{2i}\right ) ^{n}

then use binomial theorem to expand and integrate term by term. similarly for cos(x)

other than that, I can't think of any way that De Moivre's Theorem might help... it might if you got bounds on the integral.
 
You can use Identites such as \cos^4 x = \frac{\cos 4x + 4\cos 2x +3}{8} which can be derived from expanding the LHS of the theorem, to make the integrals easy.
 
Gib Z said:
You can use Identites such as \cos^4 x = \frac{\cos 4x + 4\cos 2x +3}{8} which can be derived from expanding the LHS of the theorem, to make the integrals easy.

Oh. Yeah. Thanks. And I think it represents a unit circle in the argand plane. Or more correctly, an n sided polygon which becomes a circle as n tends to infinity. You can find the nth roots of unity which would be the angle made by the adjacent vertices on the center of the polygon.
 
Gib Z said:
De Moivre's Theorem:

(\cos x + i \sin x)^n = \cos nx + i \sin nx

Proof by induction or Euler's Formula.

Explain it? There it is, no intuitive feel or understanding about it, not unless you're Gauss.

Really? I beg to differ.

Raising a complex number of modulus 1 to the n'th power multiplies the argument by n. It is saying that multiplication of (unit) complex numbers gives rotations of the complex plane.
 
matt grime said:
Really? I beg to differ.

Raising a complex number of modulus 1 to the n'th power multiplies the argument by n. It is saying that multiplication of (unit) complex numbers gives rotations of the complex plane.

And, understanding this makes finding the nth roots of a number in the complex plane pretty simple. (DeMoivre's theorem backwards, I don't know that it has a name attributed to it other than 'finding the roots')

I'm also interested in how DeMoivre's theorem can be applied to simplify finding the integrals of the type that chaoseverlasting listed in the original post. I teach DeMoivre's theorem in pre-calculus because of the beautiful simplicity of it (and the symmetry of roots of complex numbers). It'd be nice to show students that it can also be applied in calculus to trig integrals.
 
matt grime said:
Really? I beg to differ.

Raising a complex number of modulus 1 to the n'th power multiplies the argument by n. It is saying that multiplication of (unit) complex numbers gives rotations of the complex plane.

Now that you state it, it seems obvious >.< I really need to take more time to read things..we'll yea, the bit where I used Gauss, I meant the related Formula, e^{ix}=\cos x + i \sin x, he said if that wasn't immediately obviously to someone they would never be a first class mathematician.
 

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