I did probably a mistake in calculation but I cannot find it. I start from $$|c_l|^2\int_{0}^{2\pi}d\phi\int_{0}^{\pi}d\theta \sin(\theta)(\sin(\theta))^{2l}=1$$ $$I_{l}=\int_{0}^{\pi}d\theta \sin(\theta)(\sin(\theta))^{2l}=\frac{\Gamma(\frac{1}{2})\Gamma(l+1)}{\Gamma(l+\frac{3}{2})}=\frac{\sqrt{\pi} \,\,l!}{(l+\frac{1}{2})\Gamma(l+\frac{1}{2})}$$ replacing: $$\frac{2\pi|c_l|^2\sqrt{\pi} \, l!}{(l+\frac{1}{2})\Gamma(l+\frac{1}{2})}=1$$ $$|c_l|=\frac{1}{2^l \, l!}\sqrt{\frac{(l+\frac{1}{2})(2l)!}{2\pi}}$$ but the correct result is: $$\boxed{|c_l|=\frac{1}{2^l \, l!}\sqrt{\frac{(2l+1)!}{4\pi}}}$$
Someone can tell me where I was wrong and show me how to get the correct result.
Both results are the same answer...
$$|c_l|=\frac{1}{2^l \, l!}\sqrt{\frac{(2l+1)!}{4\pi}}=\frac{1}{2^l \, l!}\sqrt{\frac{(2l)!(2l+1)}{4\pi}}=\frac{1}{2^l \, l!}\sqrt{\frac{(2l)!(l+1/2)}{2\pi}}$$