Integrating sin raised to the nth power

We shall take a journey through the mind of English mathematician John Wallis and see how he in 1656 discovered intrinsic formula known as Wallis's Product. Hope you like it!

Note: I had to make this a 4-part video without sacrificing the mathematical rigour. I hope you can watch all of them for a complete proof.

Check out www.gaussianmath.com for a more indepth look and other calculus related topics.

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• wallis
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• donny
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• math

[workingballer]

you the man dony. how old are you and what college did you attend? thanks!

[donylee]

Hey,

I'm 22 years old. The college I attend to is confidential at the moment.

Thanks for your interest in my videos.

[2terence2jones]

i love calculus.

[esgerli]

Thanks a lot! You are very smart Dony!

[paulojunior201]

John Wallis was English by the way mate.

[Prey2Hunter]

Yes, that seems much easier. :)

[acidflask]

You're right, I got the formula wrong.

What I should have said was to substitute the identity

sin(x) = exp(ix) - exp(-ix) / 2 i

and use the binomial theorem to expand it out, then integrate termwise.

[Prey2Hunter]

After finding a question on yahoo answers, I started to try and figure this one out. It took me about an hour, but I achieved in getting the same answer. I learned Calculus when I was 12 (16 now). I've been addicted ever since.

[Prey2Hunter]

Im[e^inx]

=Im[cos(nx)+isin(nx)]

=sin(nx)

Integrating this isn't the same as integrating (sin(x))^n. Or is there something I am missing?

[acidflask]

Very cool. Try doing it with complex numbers; the derivation is MUCH simpler.

$I_n = \int \sin nx = \Im \int e^{i n x} = ...$

The analyticity of sin, cos and exp guarantee that doing it is OK.

[boddahboy]

I think Wallis was actually English, not American. Cool video though.