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fairandsquare · Mar 24, 2010, 10:40 AM

How can I analytically integrate cos(x)^n where n is real, but n > 1? I know the answer when n is an integer, but the method doesn't seems to apply to real numbers.

Cheers

Unknown008 · Mar 24, 2010, 11:37 AM

Well, I don't know how to do it, but from wolframalpha, it's possible...

http://www.wolframalpha.com/input/?i=int+cos(x)^n

See if you can make head or tail of it...

Maybe galactus/ebaines can provide the method to solve this?

fairandsquare · Mar 24, 2010, 12:43 PM

Thanks for your help! The hypergeometric function is a bit of a mystery to me as well! I was hoping there would be another method that didn't invole that, but if not, could anyone explain why it is required.

Thanks

galactus · Mar 25, 2010, 08:52 AM

You can use integration by parts to find the reduction formula:

$\int cos^{n}(x)dx=\frac{cos^{n-1}(x)sin(x)}{n}+\frac{n-1}{n}\int cos^{n-2}(x)dx$

Is this what you mean?

fairandsquare · Mar 25, 2010, 05:35 PM

Hi,
Thanks for your response. I believe the redustion formula only works when n is an integer. However, I don't think this equation works when n is real, as at some point if say n = 3.514, then at some point we will have to integrate cos(x)^1.514, which, if I am correct, the above equation will not solve.

Cheers

galactus · Mar 26, 2010, 03:29 PM

I don't believe there is a catchall closed form for all n.

Here is a fun one though involving Wallis' formula:

$\int_{0}^{\frac{\pi}{2}}cos^{n}(x)dx=\frac{2\cdot 4 \cdot 6\cdot\cdot\cdot (n-1)}{3\cdot 5\cdot 7\cdot \cdot \cdot n}$

for $n \;\ \text{odd}$ and $n\geq 3$

For n even and greater than or equal to 2:

$\int_{0}^{\frac{\pi}{2}}cos^{n}(x)dx=\frac{\pi}{2} \cdot \frac{1\cdot 3\cdot 5\cdot \cdot \cdot (n-1)}{2\cdot 4\cdot 6\cdot \cdot \cdot n}$

fairandsquare · Mar 29, 2010, 05:45 PM

Hi,

Thanks for the info. Those two integrals looks interesting, thanks for pointing them out.

Cheers