Integral: Moments of the semicircle distribution How can I show that $$m_{2k} = \frac{2\cdot 2^{2k}}{\pi}\int_{-\pi/2}^{\pi/2} \sin^{2k}(\theta)\cos^{2}(\theta)d\theta = \frac{2\cdot 2^{2k}}{\pi}\int_{-\pi/2}^{\pi/2} \sin^{2k}(\theta)d\theta - (2k+1)m_{2k}$$ ...? I tried writing $\cos^{2}(\theta)=1-\sin^{2}(\theta)$ to separate the integral; not sure how to get the second term.
 A: The moments of the semi-circle distribution are given by the Catalan numbers.  We can easily check this result by calculating the moments from the known probability distribution
\begin{equation}
m_{2k}=\frac{1}{\pi}\int_{-2}^{2}\lambda^{2k} \sqrt{4-\lambda^2} d\lambda
\end{equation}
where Wigner's semi-circle distribution is given by
$$
\mathcal{P}(\lambda)=\frac{1}{\pi}\sqrt{4-\lambda^2}
$$
If we change integration variables and let $\lambda$=$\sin\theta$, we obtain a trivial one-dimensional integral
\begin{eqnarray}
m_{2k}=\frac{2}{\pi} \cdot 2^{2k+2} \int_{0}^{\pi/2} \sin^{2k} \theta \cos^2 \theta d\theta\\
m_{2k}=\frac{2}{\pi} \cdot 2^{2k+2} \int_{0}^{\pi/2} \big(\sin^{2k}\theta-\sin^{2k+2}\theta\big) d\theta\\
m_{2k}=\frac{2}{\pi} \cdot 2^{2k+2} \bigg [ \int_{0}^{\pi/2} \sin^{2k}\theta d\theta -\int_{0}^{\pi/2} \sin^{2k+2}\theta d\theta \bigg ] 
\end{eqnarray}
\begin{equation}
{m_{2k}= 2 \cdot \frac{1}{k+1} \binom {2k} {k} } \ \ \ \ \ \Box
\end{equation}
The sine integrals are done using standard reduction formulas.  This concludes shows that the moments of the Wigner Semi-Circle distribution are indeed the Catalan numbers.  Note, the reduction formula applied to the integrals yields
$$
\int_{0}^{\pi/2} \sin^{2k}\theta d\theta=\frac{\Gamma(k+\frac{1}{2})\sqrt{\pi}}{2\Gamma(k+1)},  \ \ \Re(k)>-\frac{1}{2}$$ $$ \int_{0}^{\pi/2} \sin^{2k+2}\theta d\theta=\frac{\Gamma(k+\frac{3}{2})\sqrt{\pi}}{2\Gamma(k+2)}, \ \ -\frac{3}{2}<\Re(k)<-1
$$
A: Using $\cos^2 = 1-\sin^2$ as you did, we only have to write 
$$∫_{-π/2}^{π/2} sin^{2k+2} θ \ \text{d}θ  $$
in terms of $m_{2k}$. The fact that we have $2k+2$ while the result has $2k+1$ hints that we need to factor out a $\sinθ$ term,
$$∫_{-π/2}^{π/2} sin^{2k+1} θ \sinθ  \ \text{d}θ  $$
Integration by parts will yield the result.
