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.
Best Answer
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 $y=2x$, $t=\sin y$, $t=\cos^2 z$.