hard integral: $\sin(x)/\sqrt{1+2\sin(2x)}$ between $0$ and $\pi/2$

Question

I found this integral in one of my math textbooks. $$\int_0^{\pi/2}\frac{\sin(x)}{\sqrt{1+2\sin(2x)}}\mathrm{d}x$$

I tried to solve it using the substitution $u=\tan(\frac{x}{2})$ but it got me nowhere.

Answer 1

Noting that $$\int_0^{\frac{\pi}{2}} \frac{\sin x}{\sqrt{1+2 \sin (2 x)}} d x \stackrel{x\mapsto\frac{\pi}{2}-x}{=} \int_0^{\frac{\pi}{2}} \frac{\cos x}{\sqrt{1+2 \sin (2 x)}} d x$$ Averaging them gives $$ \begin{aligned} \int_0^{\frac{\pi}{2}} \frac{\sin x}{\sqrt{1+2 \sin (2 x)}} d x &=\frac{1}{2} \int_0^{\frac{\pi}{2}} \frac{\sin x+\cos x}{\sqrt{1+2 \sin (2 x)}} d x \\& =\frac{1}{2} \int_0^{\frac{\pi}{2}} \frac{d\left(\sin x-\cos x\right)}{\sqrt{3-2(\sin x-\cos x)^2}} \\ & =\frac{1}{2 \sqrt{2}}\left[\sin ^{-1}\left(\frac{\sqrt 2 (\sin x-\cos x)}{\sqrt{3}}\right)\right]_0^{\frac{\pi}{2}} \\ & =\frac{1}{\sqrt{2}}\sin ^{-1} \sqrt{\frac{2}{3}} \end{aligned} $$

Answer 2

Let $I$ denote the integral. If we make the substitution $t=x-\frac{\pi}{4}$, and notice that $\sin\left(t+\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}\left(\cos t+\sin t\right)$, as well as $\sin\left(2t+\frac{\pi}{2}\right)=\cos2t$, then the integral becomes

$$I=\underbrace{\frac{1}{\sqrt{2}}\int_{-\pi/4}^{\pi/4}\frac{\sin t}{\sqrt{1+2\cos2t}}~\mathrm{d}t}_{=0\text{ as integrand is odd}}+\frac{1}{\sqrt{2}}\int_{-\pi/4}^{\pi/4}\frac{\cos t}{\sqrt{1+2\cos2t}}~\mathrm{d}t=\frac{1}{\sqrt{2}}\int_{-\pi/4}^{\pi/4}\frac{\cos t}{\sqrt{1+2\cos2t}}~\mathrm{d}t.$$

Now, as $\cos2t=1-2\sin^2t$, we make the subsitution $u=\sin t$, so that the integral becomes

$$I=\frac{1}{\sqrt{2}}\int_{-\frac{1}{\sqrt{2}}}^\frac{1}{\sqrt{2}}\frac{\mathrm{d}u}{\sqrt{3-4u^2}}=\frac{1}{\sqrt{6}}\int_{-\frac{1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}}\frac{\mathrm{d}u}{\sqrt{1-\left(\frac{2}{\sqrt{3}}u\right)^2}}\underbrace{=}_{\text{even integrand}}\sqrt{\frac{2}{3}}\int_0^\frac{1}{\sqrt{2}}\frac{\mathrm{d}u}{\sqrt{1-\left(\sqrt{\frac{2}{3}}u\right)^2}}.$$

With a final substitution $\varphi=\frac{2}{\sqrt{3}}u$, we get that

$$I=\frac{1}{\sqrt{2}}\int_{0}^{\sqrt{\frac{2}{3}}}\frac{\mathrm{d}\varphi}{\sqrt{1-\varphi^2}}=\frac{1}{\sqrt{2}}\biggl[\arcsin\varphi\biggr]_{0}^{\sqrt{\frac{2}{3}}}=\frac{1}{\sqrt{2}}\arcsin\sqrt{\frac{2}{3}}.$$