Question

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?

Answer 1

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

Answer 2

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

Answer 3

actually i think thats wrong sry ill look it up

Answer 4

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

Answer 5

but how do you get to that?

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

You there?

Answer 11

yea chill im just trying to figure it out

Answer 12

Moo is using De Moivre's Theorem

Answer 13

Exactly!

Answer 14

Then.....?

Answer 15

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

Answer 16

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

Answer 17

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)

Answer 18

Whats your major Moo?

Answer 19

english

Answer 20

contemporary american drama

Answer 21

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

Answer 22

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

Answer 23

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

Answer 24

calculus is legit

Answer 25

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

Answer 26

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

Answer 27

Thanks a lot guys.