SOLUTION: find : int ((cos (x)))^(n) dx from 0 to pi

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Question 1208838: find : int ((cos (x)))^(n) dx from 0 to pi
Found 2 solutions by Edwin McCravy, ikleyn:
Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!

I won't do this for you, but I'll tell you what I would do.

I would use these two definite integrals:

int%28cos%5E0%28x%29%2Cdx%2C0%2Cpi%29%22%22=%22%22int%281%2Cdx%2C0%2Cpi%29%22%22=%22%22xmatrix%283%2C2%2C%22%7C%22%2Cpi%2C%22%7C%22%2C%22%22%2C%22%7C%22%2C0%29%22%22=%22%22pi-0%22%22=%22%22pi

int%28cos%5E1%28x%29%2Cdx%2C0%2Cpi%29%22%22=%22%22int%28sin%28x%29%2Cdx%2C0%2Cpi%29%22%22=%22%22sin%28x%29matrix%283%2C2%2C%22%7C%22%2Cpi%2C%22%7C%22%2C%22%22%2C%22%7C%22%2C0%29%22%22=%22%22sin%28pi%29-sin%280%29%22%22=%22%220-0%22%22=%22%220

together with this recursion formula from a table of integrals:

int%28cos%5E%28n%29%28x%29%2Cdx%2C0%2Cpi%29%22%22=%22%22matrix%283%2C2%2C%22%7C%22%2Cpi%2C%22%7C%22%2C%22%22%2C%22%7C%22%2C0%29 

Then find 

int%28cos%5E2%28x%29%2Cdx%2C0%2Cpi%29, int%28cos%5E3%28x%29%2Cdx%2C0%2Cpi%29, int%28cos%5E4%28x%29%2Cdx%2C0%2Cpi%29, etc.,

and see if you can find a pattern to the sequence of integrals, and then what
the integrals are approaching as n approaches infinity,

int%28cos%5E%28n%29%28x%29%2Cdx%2C0%2Cpi%29

Edwin

Answer by ikleyn(52754) About Me  (Show Source):
You can put this solution on YOUR website!
.

I'd like to add my two cents to it . . .


Notice that in the interval  [0,pi],  function  cos(x)  is anti-symmetric relative point  x= pi%2F2.

It implies that all integrals of  cos^n(x)dx  from  0  to  pi  are zero for all odd positive integer values of  n.