An identity involving the Bessel function of the first kind $J_0$ Today, computing $\int_{0}^{\pi/2}\sin^2(\sin^2 x)\,dx$, I found an interesting identity:
$$\sum_{k\geq 0}\frac{(-1)^k(4k)!}{(2k)!^3 4^k}=\cos(1)\cdot\sum_{k\geq 0}\frac{(-1)^k}{k!^2 4^k}.$$
How would you prove it?
 A: This answer is due to Giulio Francot, a friend of mine. His solution is extremely close to my original manipulations.

We have:
\begin{eqnarray*}
\int\limits_{0}^{\frac{\pi }{2}}\sin ^{2}(\sin ^{2}(x))dx 
&=&\int\limits_{0}^{\frac{\pi }{2}}\frac{1-\cos (2\sin ^{2}(x))}{2}dx \\
&=&\frac{
\pi }{4}-\frac{1}{2}\int\limits_{0}^{\frac{\pi }{2}}\sum\limits_{k\geq 0}
\frac{(-1)^{k}4^{k}}{(2k)!}\sin ^{4k}(x)dx \\
&=&\frac{\pi }{4}-\frac{1}{2}\sum\limits_{k\geq 0}\frac{(-1)^{k}4^{k}}{(2k)!
}\int\limits_{0}^{\frac{\pi }{2}}\sin ^{4k}(x)dx \\ 
&=&\frac{\pi }{4}-\frac{1}{2}
\sum\limits_{k\geq 0}\frac{(-1)^{k}4^{k}}{(2k)!}\frac{(4k)!}{
(4^{k}(2k)!)^{2}}\frac{\pi }{2} \\
&=&\frac{\pi }{4}-\frac{\pi }{4}\sum\limits_{k\geq 0}\frac{(-1)^{k}}{
((2k)!)^{3}}\frac{(4k)!}{4^{k}}.\tag{1}
\end{eqnarray*}
On the other hand:
\begin{eqnarray*}
\int\limits_{0}^{\frac{\pi }{2}}\sin ^{2}(\sin ^{2}(x))dx
&=&\int\limits_{0}^{\frac{\pi }{2}}\frac{1-\cos (2\sin ^{2}(x))}{2}
dx=\int\limits_{0}^{\frac{\pi }{2}}\frac{1-\cos (1-\cos (2x))}{2}dx \\
&=&\frac{\pi }{4}-\int\limits_{0}^{\frac{\pi }{2}}\frac{\cos (1)\cos (\cos
(2x))}{2}dx-\int\limits_{0}^{\frac{\pi }{2}}\frac{\sin (1)\sin (\cos (2x))}{
2}dx \\
&=&\frac{\pi }{4}-\frac{\cos (1)}{2}\int\limits_{0}^{\frac{\pi }{2}}\cos
(\cos (2x))dx=\frac{\pi }{4}-\frac{\cos (1)}{4}\int\limits_{0}^{\pi }\cos
(\cos (x))dx \\
&=&\frac{\pi }{4}-\frac{\cos (1)\pi }{4}J_{0}(1).\tag{2}
\end{eqnarray*}
A: Another approach can be this. We have $$\cos\left(x\right)J_{0}\left(x\right)=\sum_{k\geq0}\frac{\left(-1\right)^{k}}{\left(2k\right)!}x^{2k}\sum_{k\geq0}\frac{\left(-1\right)^{k}}{4^{k}k!^{2}}x^{2k}
 $$ and by Cauchy product we get $$=\sum_{k\geq0}c_{k}\left(x\right)
 $$ where $$c_{k}\left(x\right)=\left(-1\right)^{k}x^{2k}\sum_{j=0}^{k}\frac{4^{-j}}{j!^{2}\left(2k-2j\right)!}.
 $$ Now observe that $$\sum_{j=0}^{k}\frac{4^{-j}}{j!^{2}\left(2k-2j\right)!}=\frac{1}{\left(2k\right)!}\sum_{j=0}^{k}\frac{\left(2k\right)!}{j!\left(2k-2j\right)!}\frac{4^{-j}}{j!}=$$ $$=\frac{1}{\left(2k\right)!}\sum_{j=0}^{k}\frac{\left(2k\right)\left(2k-2\right)\cdots\left(2k-2j+2\right)\left(2k-1\right)\cdots\left(2k-2j+1\right)}{j!}\frac{4^{-j}}{j!}=$$ $$=\frac{1}{\left(2k\right)!}\sum_{j=0}^{\infty}\frac{k\left(k-1\right)\cdots\left(k-j+1\right)\left(k-1/2\right)\cdots\left(k-j+1/2\right)}{j!}\frac{1}{j!}=
 $$ $$=\frac{1}{\left(2k\right)!}\sum_{j=0}^{\infty}\frac{\left(-k\right)_{j}\left(-k+1/2\right)_{j}}{j!}\frac{1}{j!}=\frac{1}{\left(2k\right)!}\,_{2}F_{1}\left(-k,-k+1/2;1;1\right)
 $$ where $\left(a\right)_{j}
 $ is the Pochhammer symbol. Then, using the identity $$\,_{2}F_{1}\left(a,b;c;1\right)=\frac{\Gamma\left(c\right)\Gamma\left(c-a-b\right)}{\Gamma\left(c-a\right)\Gamma\left(c-b\right)}
 $$ we have $$\sum_{j=0}^{k}\frac{4^{-j}}{j!^{2}\left(2k-2j\right)!}=\frac{1}{\left(2k\right)!}\frac{\Gamma\left(2k+1/2\right)}{\Gamma\left(k+1\right)\Gamma\left(k+1/2\right)}=\frac{4^{k}\Gamma\left(2k+1/2\right)}{\sqrt{\pi}\left(2k\right)!^{2}}=\frac{\left(4k\right)!}{4^{k}\left(2k\right)!^{3}}
 $$ by Gamma duplication formula. So $$\cos\left(x\right)J_{0}\left(x\right)=\sum_{k\geq0}\frac{\left(-1\right)^{k}\left(4k\right)!}{4^{k}\left(2k\right)!^{3}}x^{2k}
 $$ then your identity with $x=1
 $.
P.s. I would like to thank Elajan for the inspiration.
