How to compute a $\sum_{n=0}^{\infty}{\frac{2^n}{(2n+1){2n\choose n}}}$

Your final expression has a small error. The equality you intended to write is

$$\sum_{n=0}^\infty \frac{2^n}{(2n+1)\binom{2n}{n}}=\sum_{n=0}^\infty \frac1{2^n}\int_0^{\pi/2}\sin^{2n+1}(x)\,dx$$

Now, if we change the order of the summation and integration (valid by uniform convergence), then we find that

$$\sum_{n=0}^\infty \frac{2^n}{(2n+1)\binom{2n}{n}}=\int_0^{\pi/2}\sin(x)\sum_{n=0}^\infty \frac1{2^n}\left(\sin^{2}(x)\right)^n\,dx$$

Next, sum the geometric series and carry out the resulting integral. Can you wrap this up now?

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[5px,#ffd]{\sum_{n = 0}^{\infty}{2^{n} \over \pars{2n + 1}{2n \choose n}}} = \sum_{n = 0}^{\infty}2^{n}\,{\Gamma\pars{n + 1}\Gamma\pars{n + 1} \over \Gamma\pars{2n + 2}} \\[5mm] = &\ \sum_{n = 0}^{\infty}2^{n}\int_{0}^{1} x^{n}\pars{1 - x}^{n}\,\dd x = \int_{0}^{1}\sum_{n = 0}^{\infty} \bracks{2x\pars{1 - x}}^{\, n}\,\dd x \\[5mm] = &\ \int_{0}^{1}{\dd x \over 1 - 2x\pars{1 - x}} = {1 \over 2}\int_{0}^{1}{\dd x \over x^{2} - x + 1/2} \\[5mm] = &\ {1 \over 2}\int_{0}^{1}{\dd x \over \pars{x - 1/2}^{\, 2} + 1/4} = {1 \over 2}\int_{-1/2}^{1/2}{\dd x \over x^{2} + 1/4} \\[5mm] = &\ \int_{0}^{1/2}{\dd x \over x^{2} + 1/4} = 4\,{1 \over 2}\int_{0}^{1/2}{2\,\dd x \over \pars{2x}^{2} + 1} \\[5mm] = &\ 2\int_{0}^{1}{\dd x \over x^{2} + 1} = \bbx{\pi \over 2} \\ & \end{align}