Proof for odd powers of sin: $\int(\sin x)^{2n+1}\,dx$ Prove that $\int_{0}^{\pi/2}(\sin x)^{2n+1}\,dx=\prod_{j=1}^n \frac{2j}{2j + 1}$ for odd powers of sin.
I have been able to reduce this to: $\frac{2n}{2n+1}\int_{0}^{\pi/2}(\sin x)^{2n-1}\,dx$
and I am not quite sure how to integrate $(\sin x)^{2n-1}$.
Would integration by parts be enough to solve this? If so, then how?
My implementation was to set $u = (\sin x)^{2n-1}$ and $dv = dx$ but I get a greater mess when I integrate by parts.
 A: I would say that you've already solved most of the problem yourself: your integration by parts is all the calculus you need to do.
You say "I am not quite sure how to integrate $\int (\sin x)^{2n-1}$."  So what I think you are missing is the idea that you can integrate by parts repeatedly, reducing the exponent by $2$ each time, until you get down to $\int_0^{\pi/2} \sin x dx = 1$.  (Calling the power "$2n-1$" does not make for a truly different integral than if the power is called "$2n+1$".)  This concept generally goes under the name reduction formula: see the linked wikipedia article for an introduction to how that works.
To get a sense of what's going on, I would recommend that you take a particular value of $n$ and repeatedly integrate by parts and see what you get.  E.g. if $n = 2$ then
$\int_0^{\frac{\pi}{2}} (\sin x)^5 dx = \left(\frac{2(2)}{2(2)+1}\right) \int_0^{\frac{\pi}{2}} (\sin x)^3 dx = \left(\frac{2(2)}{2(2)+1}\right) \left(\frac{2(1)}{2(1)+1)} \right) \int_0^{\frac{\pi}{2}} (\sin x) dx = \frac{4}{5} \cdot \frac{2}{3} \cdot 1 = \frac{8}{15}$.
If you do this for a few values of $n$ you will see that you are getting a product of certain fractions which is exactly equal to what you want.  (And it helps a lot to have the right hand side written down correctly!  Hence my comments above.)
You can also think of this in terms of finding an explicit formula for a recursively defined sequence if you want...but if you were comfortable with thinking about it that way I think you wouldn't have asked the question in the first place.
By some cosmic coincidence, I talked about reduction formulas in my calculus class yesterday.  A very similar reduction formula leads to an interesting formula of Wallis giving an expression for $\pi$: see here if you like.
A: PS. You can do this by integration by parts if you really, really want. Extract a $\sin^2 x$, turn it into $(1-\cos^2 x)$, and you get (if we let $I_{k} = \int_0^{\pi/2} (\sin x)^k dx$)
$I_{2n+1} = I_{2n-1} - \int_0^{\pi/2} [\sin^{2n-1}(x) \cos (x)] \cos x$.
Use integration by parts now ($u=\cos x$). You actually get a boundary term (evaluates to 0) and a multiple of the original integral, and
$\frac{2n+1}{2n} I_{2n+1} = I_{2n-1}$. 
and then some recursive fun.
A: Hint: It is a straightforward application of the beta function.
