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?

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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?

Mathematics
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\[\int\limits_{0}^{\Pi}\sin ^{2n}(\theta)d \theta\]
2n+1 is the exponent and the denomater is 2n+1
actually i think thats wrong sry ill look it up

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The answer is \[\Pi(2n)!/((2^{2n})\times (n!)^{2})\]
but how do you get to that?
So I know that e^(i*pi/2)= i
and e^(i(theta)) = cos(theta)+ isin(theta)
also cos(theta)= (e^(itheta)+e^(-itheta))/2 and sin(theta)=(e^(itheta)-e^(-itheta))/(2i)
There is also an identity, 2cos(theta)e^(itheta)=1+e^(i2theta)
You there?
yea chill im just trying to figure it out
Moo is using De Moivre's Theorem
Exactly!
Then.....?
also not to mention, \[\int\limits_{0}^{\Pi}\cos \theta ^{n}\cos n \theta d \theta=\Pi/2^{n}, n =0\]
n=0, 1, 2, 3....
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)
Whats your major Moo?
english
contemporary american drama
you know an awful lot about calculus for an English major..... lol
read Oleanna, Third, ah...Shape of things and such. You know calculus is really fun
so rewrite the integrand as sin(theta)*sin(theta) both raised to the power n?
calculus is legit
also is cos(n*theta)=cos(theta)^n? Yeah, it is..lol.
oh, bedtime. my mom is calling me downstairs with her awful hammer thingie.
Thanks a lot guys.

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