# Recursive Integral Simplification

1. Jun 11, 2004

### Nihilist Comedian

[TEX]\int{\sin^{n}(x)\cos^{m}(x)dx}[/TEX]
[TEX]=\frac{\sin^{n+1}(x)\cos^{m-1}(x)}{n+1}+\frac{m-1}{n+1}\int{\sin^{n+2}(x)\cos^{m-2}(x)dx}[/TEX]

That was quite easy, but it's the simplification process following this that throws me. My answer is perfectly correct, but it is simplified in the answers (in my maths book) to the following form.

[TEX]\frac{\sin^{n+1}(x)\cos^{m-1}(x)}{n+m}+\frac{m-1}{n+m}\int{\sin^{n}(x)\cos^{m-2}(x)dx}[/TEX]

The form that I had it in can be used to calculate integrals for specific values of n an m, though in an exam, I believe that I'd have to express it in simpler form to get full marks.

Thanks for the help.

Last edited: Jun 14, 2004
2. Jun 14, 2004

### Nihilist Comedian

Sorry, forgot to put in the "n" and "m" in the original function.

Any help now?

3. Jun 14, 2004

### arildno

You have the following identity:
$$\int\sin^{n+2}x\cos^{m-2}xdx=\int\sin^{n}x(1-\cos^{2}x)\cos^{m-2}xdx$$

4. Jun 15, 2004

### Nihilist Comedian

Thanks. It's really quite simple. I can't believe I missed that!