Solving the integral :$\int_0^{\frac{\pi}{2}}\frac{\sin x}{1+\sqrt{\sin2 x}}dx$ Solving the integral :
$$\int_0^{\frac{\pi}{2}}\frac{\sin x}{1+\sqrt{\sin 2x}}dx=?$$

My try:
$$\int_0^{\frac{\pi}{2}}\frac{\sin x}{1+\sqrt{\sin 2x}}\cdot \frac{1-\sqrt{\sin 2x}}{1-\sqrt{\sin 2x}}dx$$
$$\int_0^{\frac{\pi}{2}}\frac{\sin x(1-\sqrt{\sin 2x})}{1-\sin 2x}dx$$
$$(\sin x-\cos x)^2=1-\sin 2x$$
So :
$$\int_0^{\frac{\pi}{2}}\frac{\sin x(1-\sqrt{\sin 2x})}{(\sin x-\cos x)^2}dx$$
Now what?
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 A: The given integral equals
$$ \frac{1}{2}\int_{0}^{\pi}\frac{\sin\tfrac{x}{2}}{1+\sqrt{\sin x}}\,dx =\frac{1}{2\sqrt{2}}\int_{0}^{\pi}\frac{\sqrt{1-\cos x}}{1+\sqrt{\sin x}}\\=\frac{1}{2\sqrt{2}}\int_{0}^{\pi/2}\frac{\sqrt{1-\cos x}+\sqrt{1+\cos x}}{1+\sqrt{\sin x}}\,dx\\
=\frac{1}{2}\int_{0}^{\pi/2}\frac{\sqrt{1+\sin x}}{1+\sqrt{\sin x}}\,dx$$
or
$$ \frac{1}{2}\int_{0}^{1}\frac{\sqrt{1+x}}{\sqrt{1-x^2}(1+\sqrt{x})}\,dx = \frac{1}{2}\int_{0}^{1}\frac{dx}{(1+\sqrt{x})\sqrt{1-x}}=\int_{0}^{1}\frac{x\,dx}{(1+x)\sqrt{1-x^2}}$$
which equals $\frac{\pi}{2}-\int_{0}^{1}\frac{dx}{(1+x)\sqrt{1-x^2}}$. On the other hand
$$ \int_{0}^{1}\frac{dx}{(1+x)\sqrt{1-x^2}}=\int_{0}^{\pi/2}\frac{d\theta}{1+\sin\theta}=\int_{0}^{\pi/2}\frac{d\theta}{1+\cos\theta}=\int_{0}^{\pi/2}\frac{d\theta}{2\cos^2\tfrac{\theta}{2}}=1, $$
hence:
$$\boxed{\int_{0}^{\pi/2}\frac{\sin x}{1+\sqrt{\sin(2x)}}\,dx = \color{blue}{\frac{\pi}{2}-1},} $$
no particular elliptic integral like $K\left(\frac{1}{2}\right)$ or $E\left(\frac{1}{2}\right)$ is really involved.
