Integral of the day

Integral involving the compostion of sin function with itself

We prove the following integral: $$\int_{0}^{\frac{\pi}{2}} \sin\sin xdx = 1-\frac{1}{3!!^2}+\frac{1}{5!!^2}-\frac{1}{7!!^{2}}+...$$ Proof
Recall the series expansion of $\sin(x)$ $$\sin x = \sum_{n=0}^{\infty } \frac{(-1)^n}{(2n+1)!}x^{2n+1} \quad \forall x\in \mathbb{R}$$ Therefore $$\begin{align*} \int_{0}^{\frac{\pi}{2}} \sin\sin xdx =& \int_{0}^{\frac{\pi}{2}} \sum_{n=0}^{\infty } \frac{(-1)^n}{(2n+1)!}\sin^{2n+1}(x)dx\\ =& \sum_{n=0}^{\infty } \frac{(-1)^n}{(2n+1)!} \int_{0}^{\frac{\pi}{2}}\sin^{2n+1}(x)dx\\ =& \sum_{n=0}^{\infty } \frac{(-1)^n}{(2n+1)!} \int_{0}^{1}w^{2n+1}(1-w^2)^{-\frac{1}{2}}dw \quad (w\mapsto \sin x)\\ =& \sum_{n=0}^{\infty } \frac{(-1)^n}{(2n+1)!} \int_{0}^{1} \frac{t^{n}(1-t)^{-\frac{1}{2}}}{2}dt \quad (t \mapsto w^2)\\ =&\sum_{n=0}^{\infty } \frac{(-1)^n}{(2n+1)!} \frac{B(n+1,\frac{1}{2})}{2}\\ =&\sum_{n=0}^{\infty } \frac{(-1)^n}{(2n+1)!} \frac{\Gamma(n+1)\Gamma(\frac{1}{2})}{2\Gamma(n+1+\frac{1}{2})}\\ =&\sum_{n=0}^{\infty } \frac{(-1)^n}{(2n+1)!} \frac{n! \sqrt{\pi}}{2\Gamma(n+1+\frac{1}{2})} \quad (\Gamma(\frac{1}{2}) = \sqrt{\pi}))\\ \end{align*}$$ Now recall the Legrende duplication formula $$\sqrt{\pi} \Gamma(2z) = 2^{2z-1} \Gamma(z)\Gamma(z+\frac{1}{2})$$ Applying the formula to $\Gamma(n+1+\frac{1}{2})$ $$\begin{align*} \sum_{n=0}^{\infty } \frac{(-1)^n}{(2n+1)!} \frac{n! \sqrt{\pi}}{2\Gamma(n+1+\frac{1}{2})} = &\sum_{n=0}^{\infty } \frac{(-1)^n}{(2n+1)!} \frac{2^{2n}n! \Gamma(n+1)}{\Gamma(2n+2)} \\ =&\sum_{n=0}^{\infty } \frac{(-1)^n}{(2n+1)!} \frac{2^{2n}n!^2 }{(2n+1)!}\\ =&\sum_{n=0}^{\infty } (-1)^n \frac{2^{2n}n!^2 }{(2n+1)!^2}\\ =&\sum_{n=0}^{\infty } (-1)^n \frac{1 }{(2n+1)!!^2} \end{align*}$$ $$\boxed{\int_{0}^{\frac{\pi}{2}} \sin\sin xdx = \sum_{n=0}^{\infty } (-1)^n \frac{1 }{(2n+1)!!^2}}$$