# N-dim integral

1. Sep 17, 2011

### pirce

Hello

I have problem with integrate

$$\int_{0}^{\frac{\pi}{2}}d\phi_1...\int_{0}^{\frac{\pi}{2}}d\phi_{n-2}sin^{2(n-1)}\phi_1...sin^{2(n-k)}\phi_k ...sin^4\phi_{n-2}$$

2. Sep 17, 2011

### jackmell

Ok, you got to start small with these things then build it back up. So how about starting with n=2:

$$\int_0^a\int_0^a \sin^2(\phi_1)\sin^4(\phi_2) d\phi_1 d\phi_2$$

Alright, get that one straight, then add another one, then another one. Do maybe 3,4 or five that way and you'll (hopefully) see a trend that you can then use deduction to deduce the value for the general expression.

3. Sep 17, 2011

### pirce

Using the Mathematica I found that
$$\int_{0}^{\frac{\pi}{2}}Sin^2xdx=\frac{\pi}{4}$$$$\int_{0}^{\frac{\pi}{2}}Sin^4xdx=\frac{3\pi}{16}$$
$$\int_{0}^{\frac{\pi}{2}}Sin^6xdx=\frac{5\pi}{32}$$$$\int_{0}^{\frac{\pi}{2}}Sin^8xdx=\frac{35\pi}{256}$$
$$\int_{0}^{\frac{\pi}{2}}Sin^{10}xdx=\frac{63\pi}{512}$$

so I can write

$$\int_{0}^{\frac{\pi}{2}}Sin^{2n}xdx=\frac{a_{2n} \pi }{2^{2n}}$$

but I cant find any logical expression which describes an

4. Sep 17, 2011

### jackmell

Keep in mind the variables are separated so won't the answer be some kind of product like:

$$\displaystyle\prod_{n=1}^N (I_n)$$

I think so anyway. So the $a_n$ term may be a problem. Ok, what happens when you just solve for the antiderivative when n=2, n=3, n=4, n=5. Can you see some kind of trend there?.

5. Sep 17, 2011

### jackmell

I think I may have led you astray on this and now suggest we focus on the recursive formula:

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

Haven't worked it out yet but it looks encouraging.