A curious sin-integral While contending with a certain Fourier series, I stumbled on an incredibly simple evaluation (numerically) of a slightly complicated-looking sin-integral.
So, I wish ask:

Question. Is this really true? If so, any proof?
$$I:=\int_0^{\frac{\pi}2}\frac{\sin x}{1+\sqrt{\sin 2x}}\,dx=\frac{\pi}2-1.$$

ADDED. I'm an experimentalist and I find many many results. Some I could find being discovered earlier after checking the literature. For others, either I don't find them easily or I might be tired of looking and hope someone else points them out to me. I'm mostly interested in sharing and having fun, not seeking recognition of any sort. However, one thing is for sure: I don't give oxygen to rude comments.
 A: We have 
\begin{align}
& 2\int_0^{\pi/2}\frac{\sin x}{1+\sqrt{\sin 2x}} \, dx=\int_0^{\pi/2}\frac{\sin x+\cos x}{1+\sqrt{\sin 2x}} \, dx=\frac12\int_0^\pi\frac{\sqrt{1+\sin y}}{1+\sqrt{\sin y}} \, dy \\[6pt]
= {} &\int_0^{\pi/2}\frac{\sqrt{1+\sin y}}{1+\sqrt{\sin y}} \, dy =\int_0^1\frac{\sqrt{1+t}}{(1+\sqrt{t})\sqrt{1-t^2}} \, dt=\int_0^1\frac{dt}{(1+\sqrt{t})\sqrt{1-t}} \\[6pt]
= {} &2\int_0^{\pi/2}\frac{\cos z}{1+\cos z} \, dz=\pi-2\int_0^{\pi/2}\frac1{1+\cos z} \,dz= \pi-2\tan\frac{z}2\bigg|_0^{\pi/2}=\pi-2,
\end{align}
where we used substitutions $y=2x$, $t=\sin y$, $t=\cos^2 z$.
A: 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.
A: Here is another approach:
$$\begin{align}2I&=\int_{0}^{\pi/2}\frac{\sin t+\cos t}{1+\sqrt{\sin 2t}}\,dt\\&=\int_{0}^{\pi/2}\frac{\sin t+\cos t}{1+\sqrt{1-(\sin t-\cos t)^2}}\,dt\\& =\underbrace{ \int_{-\pi/2}^{\pi/2}\frac{\cos u}{1+\cos u}\,du}_{\sin t-\cos t=\sin u }\\&=2\int_0^{\pi/2}\left(1-\dfrac{1}{1+\cos u}\right)\,du\end{align}$$ 
$$\boxed{I=\frac{\pi}{2}-1}$$
