Integrating a Complex Math Problem - Need Help!

[pirce]

Hello

I have problem with integrate

<br /> \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}<br />

Please help me.

 

[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.

 

[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 can't find any logical expression which describes an

 

[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?.

 

[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.