## anonymous 5 years ago 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?

1. anonymous

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

2. anonymous

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

3. anonymous

actually i think thats wrong sry ill look it up

4. anonymous

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

5. anonymous

but how do you get to that?

6. anonymous

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

7. anonymous

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

8. anonymous

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

9. anonymous

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

10. anonymous

You there?

11. anonymous

yea chill im just trying to figure it out

12. anonymous

Moo is using De Moivre's Theorem

13. anonymous

Exactly!

14. anonymous

Then.....?

15. anonymous

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

16. anonymous

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

17. 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)

18. anonymous

19. anonymous

english

20. anonymous

contemporary american drama

21. anonymous

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

22. anonymous

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

23. anonymous

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

24. anonymous

calculus is legit

25. anonymous

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

26. anonymous

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

27. anonymous

Thanks a lot guys.