Evaluating a sum involving binomial coefficient in denominator I came across the following sum:
$$\sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)^2}\frac{4^k}{{2k \choose k}}$$

I thought that this can be evaluated using the expansion of $\dfrac{\sin^{-1}x}{\sqrt{1-x^2}}$ but I couldn't make any use of it. Then I tried to use the following:
$$\frac{1}{(2k+1)}\frac{1}{{2k \choose k}}=\frac{\Gamma(k+1)\Gamma(k+1)}{\Gamma(2k+2)}=\int_0^1 x^k(1-x)^k\,dx$$
but this didn't help either and now I am stuck. 
Any help is greatly appreciated. Thanks!

EDIT:
The sum originated from the following definite integral:
$$\int_0^{\pi/2}\tan^{-1}(\sin x)\,dx$$
 A: Here is a little something. Along the same line as your thoughts. 
$$\sum_{k=0}^{\infty}\frac{(-1)^{k}4^{k}}{(2k+1)^{2}\binom{2k}{k}}$$
$$=\sum_{k=0}^{\infty}\frac{(-1)^{k}4^{k}k!k!}{(2k+1)(2k+1)(2k)!}$$
$$=\sum_{k=0}^{\infty}\frac{(-1)^{k}4^{k}k!k!}{(2k+1)(2k+1)!}$$
$$=\sum_{k=0}^{\infty}\frac{(-1)^{k}4^{k}\Gamma(k+1)\Gamma(k+1)}{(2k+1)\Gamma(2k+2)}$$
$$=\sum_{k=0}^{\infty}\frac{(-1)^{k}4^{k}\beta(k+1,k+1)}{2k+1}$$
$$=\sum_{k=0}^{\infty}\frac{(-1)^{k}}{2k+1}\int_{0}^{1}(4x(1-x))^{n}dx$$
$$=\sum_{k=0}^{\infty}\frac{(-1)^{k}}{2k+1}\int_{0}^{1}(1-(2x-1)^{2})^{n}dx$$
Let $t=2x-1$
$$1/2\sum_{k=0}^{\infty}\frac{(-1)^{k}}{2k+1}\int_{-1}^{1}(1-t^{2})^{n}dt$$
$$=\sum_{k=0}^{\infty}\frac{(-1)^{k}}{2k+1}\int_{0}^{1}(1-t^{2})^{n}dt$$
let $t=\sin(x)$
$$\int_{0}^{\frac{\pi}{2}}\cos^{2}(x)\sum_{k=0}^{\infty}\frac{(-1)^{k}\cos^{2k-1}}{2k+1}dx$$
Note the series is arctan at cos(x)
$$=\int_{0}^{\frac{\pi}{2}}\tan^{-1}(\cos(x))dx$$
Parts:
$u=\tan^{-1}(\cos(x)), \;\ dv=dx, \;\ du=\frac{-\sin(x)}{1+\cos^{2}(x)}dx, \;\ v=x$
$$0+\int_{0}^{\frac{\pi}{2}}\frac{x\sin(x)}{1+\cos^{2}(x)}dx$$
This is as far as I have gotten for now. It seems to me this may have been done on the site at some point if we can find it. 
See here for the evaluation of this integral.  SOS done it sometime back.
http://sos440.tistory.com/83
A: $$ \begin{align} \sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)^2}\frac{4^k}{{2k \choose k}} &= \sum_{k=0}^{\infty} \frac{(-1)^{k}}{2k+1} \int_{0}^{\pi /2} \sin^{2k+1} (x) \, dx  \\ &= \int_{0}^{\pi /2} \sum_{n=0}^{\infty} \frac{(-1)^{k} \sin^{2k+1} (x)}{2k+1} \, dx\\ &= \int_{0}^{\pi /2} \arctan (\sin x) \, dx \\ &= \int_{0}^{1} \frac{\arctan t}{\sqrt{1-t^{2}}} \, dt \end{align}$$
Let $ \displaystyle I(a) = \int_{0}^{1} \frac{\arctan (at)}{\sqrt{1-t^{2}}} \ dt$.
Then differentiating under the integral sign, 
$$ \begin{align}  I'(a) &= \int_{0}^{1} \frac{t}{(1+a^{2}t^{2})\sqrt{1-t^{2}}} \, dt \\ &= \int_{0}^{1} \frac{1}{[1+a^{2}(1-u^{2})]u} \, u \, du \\ &= \frac{1}{1+a^{2}} \int_{0}^{1} \frac{1}{1-\left( \frac{au}{\sqrt{1+a^{2}}}\right)^{2}} \, du \\ &= \frac{1}{a \sqrt{1+a^{2}}} \text{arctanh} \left( \frac{a}{\sqrt{1+a^{2}}} \right) \\ &= \frac{1}{a\sqrt{1+a^{2}}} \frac{1}{2} \ln \Big((a+\sqrt{1+a^{2}})^{2} \Big) \\ &= \frac{1}{a \sqrt{1+a^{2}}} \ln \left( a+ \sqrt{1+a^{2}} \right) \\ &= \frac{1}{a \sqrt{1+a^{2}}} \text{arcsinh}(a) . \end{align}$$
And then integrating back,
$$ \begin{align} I(1)-I(0) = I(1) &= \int_{0}^{1} \frac{\text{arcsinh}(a)}{a \sqrt{1+a^{2}}} \, da \\ &= - \text{arcsinh}(a) \text{arcsinh}(\frac{1}{a}) \Bigg|^{1}_{0} + \int_{0}^{1} \frac{\text{arcsinh}(\frac{1}{a})}{\sqrt{1+a^{2}}} \, da \\ &= - \text{arcsinh}^{2}(1) +  \int_{0}^{1} \frac{\text{arcsinh}(\frac{1}{a})}{\sqrt{1+a^{2}}} \, da \\ &= - \ln^{2}(1+\sqrt{2}) + \int_{0}^{1} \frac{\text{arcsinh}(\frac{1}{a})}{\sqrt{1+a^{2}}} \, da . \end{align}$$
Now let $ \displaystyle w = \frac{1}{a}$.
Then
$$ I(1) = - \ln^{2}(1+\sqrt{2}) + \int_{1}^{\infty} \frac{\text{arcsinh}(w)}{w \sqrt{1+w^{2}}} \, dw$$
$$ =  - \ln^{2}(1+\sqrt{2}) + I(\infty) - I(1) .$$
Therefore,
$$ \begin{align} I(1) &= - \frac{\ln^{2}(1+\sqrt{2})}{2} + \frac{I(\infty)}{2} \\ &= - \frac{\ln^{2}(1+\sqrt{2})}{2} + \frac{\pi}{4} \int_{0}^{1} \frac{1}{\sqrt{1-t^{2}}} \, dt \\ &= - \frac{\ln^{2}(1+\sqrt{2})}{2} + \frac{\pi^{2}}{8} . \end{align}$$
