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Gib Z

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[tex](\cos x + i \sin x)^n = \cos nx + i \sin nx[/tex]

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

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

3. [tex]\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[/tex], m can not equal 1, positive.

4. [tex]\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[/tex], 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.

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Gib Z

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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.

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[tex]\sin^n(x)=\left (\frac{e^{ix}+e^{-ix}}{2i}\right ) ^{n}[/tex]

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.

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Gib Z

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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.

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matt grime

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Really? I beg to differ.De Moivre's Theorem:

[tex](\cos x + i \sin x)^n = \cos nx + i \sin nx[/tex]

Proof by induction or Euler's Formula.

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

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.

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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')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.

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.

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Gib Z

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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, [tex]e^{ix}=\cos x + i \sin x[/tex], he said if that wasn't immediately obviously to someone they would never be a first class mathematician.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.

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