This is not as crisp as Fedor's solution (which I am accepting), but it might help to see alternative techniques. In fact, I very much welcome others to join the effort (for pedagogical reasons).

From the geometric series expansion, $\frac1{1+\sqrt{\sin 2x}}=\sum_{n=0}^{\infty}(-1)^n\sin^{\frac{n}2}2x$. From the Euler's beta,
\begin{align}
\int_0^{\frac{\pi}2}\sin x\,\sin^{\frac{n}2}2x\,dx=2^{\frac{n}2}\int_0^{\frac{\pi}2}\sin^{\frac{n}2+1}x\,\cos^{\frac{n}2}x\,dx=\binom{\frac{n}2+\frac12}{\frac{n}2}^{-1}. \end{align}
Therefore, we have
\begin{align}
I=\sum_{n=0}^{\infty}(-1)^n\binom{\frac{n}2+\frac12}{\frac{n}2}^{-1}
&=\sum_{n=0}^{\infty}\left[\binom{n+\frac12}n^{-1}-\binom{n+1}{n+\frac12}^{-1}\right] \\
&=\sum_{n=0}^{\infty}\left[\frac{2^{2n}}{(2n+1)\binom{2n}n}-\frac{\pi}2\frac{\binom{2n+2}{n+1}}{2^{2n+2}}\right]. \end{align}
Letting $a_n:=2^{2n}\binom{2n}n^{-1}$, we may rewrite $I=\sum_{n\geq0}\left[\frac{a_n}{2n+1}-\frac{\pi}{2a_n}\right]+\frac{\pi}2$. Stirling's approximation shows that $a_n\sim\sqrt{\pi n}$ and hence $\frac{a_n}{2n+1}-\frac{\pi}{2a_n}\sim\frac1{n^{\frac32}}$. This ensures the integral $I$ exists, despite the fact that both $\sum\frac{a_n}{2n+1}$ and $\sum\frac{\pi}{2a_n}$ diverge, individually. On the other hand, we know
$$ f(x):=\sum_{n\geq 0} \frac{a_nx^{2n}}{2n+1} =\frac{\sin^{-1}x} {x\sqrt{1-x^2}}
\qquad \text{and} \qquad
g(x):=\sum_{n\geq 0}\frac{x^{2n}}{a_n} =\frac{1}{\sqrt{1-x^2}}. $$
Invoking Abel's Theorem and using L'Hopital's Rule, compute that
$$\sum_{n\geq0}\left[\frac{a_n}{2n+1}-\frac{\pi}{2a_n}\right]=\lim_{x\to 1-}\left(f(x)-\frac{\pi}2 g(x)\right)=-1.$$
In the end, we arrive at $I=\frac{\pi}2-1$ as required.

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