# 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?



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.
$$I=\int_0^{\frac{\pi}{2}}\frac{\sin x}{1+\sqrt{\sin 2x}}\,dx$$ Using the fact that: $$\int_a^b f(x) dx=\frac{1}{2} \int_a^b (f(x)+f(a+b-x))dx$$ $$\Rightarrow I=\frac12 \int_0^\frac{\pi}{2} \frac{\sin x+\cos x}{1+\sqrt{\sin 2x}}dx$$Now let's substitute $$\,\sin x-\cos x=u\,$$ $$\rightarrow u^2=1-\sin(2x)\rightarrow\sin(2x)=1-u^2$$ $$I=\int_{0}^{1}\frac{du}{1+\sqrt{1-u^2}}$$ Letting $$u=\sin t$$ yields: $$I=\int_0^{\frac{\pi}{2}}\frac{\cos t}{1+\cos t}\,dt= \int_0^{\frac{\pi}{2}}\left(\frac{1+\cos t}{1+\cos t}-\frac{1}{1+\cos t}\right)\,dt$$ $$\Rightarrow I=\frac{\pi}{2}-\frac{1}{2}\int_0^\frac{\pi}{2}\frac{1}{\cos^2\frac{t}{2}}\,dt=\frac{\pi}{2}-1$$