I need to integrate $\displaystyle I=\int\sin^n(x)\cos^m(x)\,dx$ where $\displaystyle (n,m)\in \mathbb{R}$. How do I do that?

Printable View

- Jul 29th 2010, 11:06 AMfobos3Integrate sin^n(x)cos^m(x)
I need to integrate $\displaystyle I=\int\sin^n(x)\cos^m(x)\,dx$ where $\displaystyle (n,m)\in \mathbb{R}$. How do I do that?

- Jul 29th 2010, 11:19 AMAlso sprach Zarathustra
Here are some ideas for start...

Integrals: sin^n(x)cos^m(x), sinh^n(x)cosh^m(x) - Jul 29th 2010, 03:03 PMfobos3
The problem I'm having is $\displaystyle \exists (m,n)\notin\mathbb{N}$ so no integration by parts.

- Jul 29th 2010, 03:20 PMAlso sprach Zarathustra
But, $\displaystyle \mathbb{N}\subset \mathbb{R}$, anyway I hardly believe you find such formula, you can find close formula for $\displaystyle m,n\in \mathbb{Z}$, I think it is the top, but maybe I wrong here...

- Jul 29th 2010, 03:31 PMVlasev
For any $\displaystyle n \in \mathbb{R}$, the derivative $\displaystyle \frac{d}{dx}x^n = n x^{n-1}$. Similarly for the antiderivative. In the integration by parts you are using the above plus the chain rule so the integration should work for $\displaystyle (n,m) \in \mathbb{R}$. Of course, you are going to have to exclude some special values like (n,m) = (0,0) or (1,1), etc

Although if you try n = 2/3 and m = 3/4, Mathematica will give you something involving hypergeometric series.

If you try n = 1/2 and m = 2, you get an elliptic integral.

I did some numerical integration for values of n and m for the integration by parts identity and it all works well as long as we work in the given ranges, i.e. n, m >1. - Jul 30th 2010, 02:12 AMfobos3
Mathematica gives me $\displaystyle \int\sin^mx\cos^nx\,dx=-\,_2F_1\left(\dfrac{1-m}{2},\dfrac{1+n}{2},\dfrac{3+n}{2},\cos^2x\right) \dfrac{\sin^{1+m}x}{\left(\sqrt{\sin^2x}\right)^{1 +m}}\dfrac{1}{1+n}\cos^{1+n}x$