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Mathematics 29 Online
OpenStudy (anonymous):

how do you get a formula for an integral from 0 to pi of sine of theta raised to the power of 2n where n is an integer from 0?

OpenStudy (anonymous):

\[\int\limits_{0}^{\Pi}\sin ^{2n}(\theta)d \theta\]

OpenStudy (anonymous):

2n+1 is the exponent and the denomater is 2n+1

OpenStudy (anonymous):

actually i think thats wrong sry ill look it up

OpenStudy (anonymous):

The answer is \[\Pi(2n)!/((2^{2n})\times (n!)^{2})\]

OpenStudy (anonymous):

but how do you get to that?

OpenStudy (anonymous):

So I know that e^(i*pi/2)= i

OpenStudy (anonymous):

and e^(i(theta)) = cos(theta)+ isin(theta)

OpenStudy (anonymous):

also cos(theta)= (e^(itheta)+e^(-itheta))/2 and sin(theta)=(e^(itheta)-e^(-itheta))/(2i)

OpenStudy (anonymous):

There is also an identity, 2cos(theta)e^(itheta)=1+e^(i2theta)

OpenStudy (anonymous):

You there?

OpenStudy (anonymous):

yea chill im just trying to figure it out

OpenStudy (anonymous):

Moo is using De Moivre's Theorem

OpenStudy (anonymous):

Exactly!

OpenStudy (anonymous):

Then.....?

OpenStudy (anonymous):

also not to mention, \[\int\limits_{0}^{\Pi}\cos \theta ^{n}\cos n \theta d \theta=\Pi/2^{n}, n =0\]

OpenStudy (anonymous):

n=0, 1, 2, 3....

OpenStudy (anonymous):

this might also be a hint, but e^(i*2pi*k) where k is any integer is 1. (for reasons I fully don't know)

OpenStudy (anonymous):

Whats your major Moo?

OpenStudy (anonymous):

english

OpenStudy (anonymous):

contemporary american drama

OpenStudy (anonymous):

you know an awful lot about calculus for an English major..... lol

OpenStudy (anonymous):

read Oleanna, Third, ah...Shape of things and such. You know calculus is really fun

OpenStudy (anonymous):

so rewrite the integrand as sin(theta)*sin(theta) both raised to the power n?

OpenStudy (anonymous):

calculus is legit

OpenStudy (anonymous):

also is cos(n*theta)=cos(theta)^n? Yeah, it is..lol.

OpenStudy (anonymous):

oh, bedtime. my mom is calling me downstairs with her awful hammer thingie.

OpenStudy (anonymous):

Thanks a lot guys.

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