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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? $$I:=\int_0^{\frac{\pi}2}\frac{\sin x}{1+\sqrt{\sin 2x}}\,dx=\frac{\pi}2-1.$$

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    $\begingroup$ Substitution $y = \cos x$ yields $\int_0^1 \frac{1}{1+\sqrt{2y\sqrt{1-y^2}}}dy$. Not sure if that's easier, though. $\endgroup$ – Glorfindel Jun 16 '17 at 20:16
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    $\begingroup$ Further substituting $y = 2t/(1+t^2)$ yields a rational function of $t$ and $\sqrt{1+t^2}$, and then yet another substitution $t=(1-u^2)/2u$ makes it a rational function of $u$, thus reducing to a problem known to be solvable. $\endgroup$ – Noam D. Elkies Jun 16 '17 at 20:30
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    $\begingroup$ @NoamD.Elkies substituting $y=2t/(1+t^2)$ I get $\sqrt{t-t^3}$ involved, not $\sqrt{1+t^2}$. I doubt, really, that indefinite integral is expressed in elementary functions. $\endgroup$ – Fedor Petrov Jun 16 '17 at 21:07
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    $\begingroup$ Yes, I wondered too: the functions $u := \sin x$ and $v := \sqrt{\sin 2x}$ satisfy $v^4 = 2u^2(1-u^2)$, which seems to define a genus $1$ curve (actually a twist of $y^2=x^3-x$, up to birational equivalence). So that seems to preclude the possibility of obtaining a rational function for the integrand. $\endgroup$ – RP_ Jun 16 '17 at 21:39
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    $\begingroup$ \begin{align} \frac{1}{2} \left(\tan ^{-1}\left(1-\frac{\sqrt{2}}{\sqrt{\tan x}}\right)-\tan ^{-1}\left(\frac{\sqrt{2}}{\sqrt{\tan x}}+1\right)\right)+\frac{1}{2\sqrt{2}}\log \left(\frac{\sqrt{\tan x}+1}{\sqrt{\tan x}-1}\right)\\+\frac12\frac{1-\sqrt{\sin 2 x}}{\cos x-\sin x}-\frac{1}{\sqrt{2}}\tanh ^{-1}\left(\frac{\tan \left(\frac{x}{2}\right)+1}{\sqrt{2}}\right)+\frac{1}{4} \log \left(\frac{\sqrt{\tan x}+\sqrt{\cot x}-\sqrt{2}}{\sqrt{\tan x}+\sqrt{\cot x}+\sqrt{2}}\right) \end{align} $\endgroup$ – Nemo Jun 16 '17 at 23:19
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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 $y=2x$, $t=\sin y$, $t=\cos^2 z$.

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    $\begingroup$ That's a nice maneuvering. $\endgroup$ – T. Amdeberhan Jun 16 '17 at 21:28
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    $\begingroup$ And formatting. ( ͡° ͜ʖ ͡°) $\endgroup$ – Mateen Ulhaq Jun 18 '17 at 12:11
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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.

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    $\begingroup$ Assuming the problem is a pesky mosquito, I'd have to say Fedor used a mosquito swatter and you used a hydrogen bomb to take care of it. :D +1 $\endgroup$ – Kugelblitz Jun 17 '17 at 5:22
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Here is another approach: $$\begin{align}2I&=\int_{0}^{\pi/2}\frac{\sin t+\cos t}{1+\sqrt{\sin 2t}}\,dt\\&=\int_{0}^{\pi/2}\frac{\sin t+\cos t}{1+\sqrt{1-(\sin t-\cos t)^2}}\,dt\\& =\underbrace{ \int_{-\pi/2}^{\pi/2}\frac{\cos u}{1+\cos u}\,du}_{\sin t-\cos t=\sin u }\\&=2\int_0^{\pi/2}\left(1-\dfrac{1}{1+\cos u}\right)\,du\end{align}$$ $$\boxed{I=\frac{\pi}{2}-1}$$

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  • $\begingroup$ yeah much easier :) $\endgroup$ – Zeno San Jun 20 '17 at 5:37

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