2

This definite integral has an equivalent closed form in terms of complete elliptic integrals,

$$\begin{align*} I &= \int_0^\tfrac\pi2 \frac{\sqrt{\sin x}}{1+\cos x} \, dx \\ & = 2 - \sqrt{\frac2\pi}\, \Gamma^2\left(\frac34\right) \\ &= \boxed{2 - 2\sqrt2 \, E\left(\frac1{\sqrt2}\right) + \sqrt2 \, K\left(\frac1{\sqrt2}\right)} \tag{$*$} \end{align*}$$

Q: Is there any way to algebraically reduce or transform $I$ to more readily obtain this elliptic integral form, without leaning on beta/gamma functions? Having made a similar connection recently, I'm wondering if the same can be done here.

Working backward from $(*)$, we have

$$\begin{align*} I &= 2 - 2\sqrt2 \int_0^\tfrac\pi2 \sqrt{1-\frac12\sin^2t} \, dt + \sqrt2 \int_0^\tfrac\pi2 \frac{dt}{\sqrt{1-\frac12\sin^2t}} \, dt \\ &= 2 \left(1 - \int_0^\tfrac\pi2 \frac{\cos^2t}{\sqrt{1+\cos^2t}} \, dt\right) \end{align*}$$

I've tried replacing $1=\int_0^{\pi/2} f(t) \, dt$ but I'm not sure if there's a clever choice of $f(t)$ that will cooperate best with the other integrand. For example, the trivial candidates $f\in\{\sin,\cos\}$ give

$$I = 2 \int_0^\tfrac\pi2 \frac{\sin t \sqrt{1+\cos^2t} - \cos^2t}{\sqrt{1+\cos^2t}} \, dt = 2 \int_0^\tfrac\pi2 \frac{\cos t \sqrt{1+\cos^2t} - \cos^2t}{\sqrt{1+\cos^2t}} \, dt \tag1$$

On the other hand, substituting $\cos y = \tan \dfrac x2$ produces a similar denominator,

$$I = \sqrt2 \int_0^\tfrac\pi2 \frac{\sin y \sqrt{\cos y}}{\sqrt{1+\cos^2y}} \, dy = \sqrt2 \int_0^\tfrac\pi2 \sqrt{\cos y} \sqrt{\frac{1-\cos^2y}{1+\cos^2y}} \, dy \tag2$$

but I cannot see any way to relate either form in $(1)$ with $(2)$.

user170231
  • 25,320

3 Answers3

1

For the antiderivative $$I=\int\frac{\sqrt{\sin (x)}}{1+\cos (x)}\,dx$$ $$\sqrt{\sin (x)}=t \quad\implies\quad x=\sin ^{-1}\left(t^2\right)\quad\implies \quad dx=\frac{2 t}{\sqrt{1-t^4}}\,dt$$ $$I=\int\frac{2 t^2}{1-t^4+\sqrt{1-t^4}}\,dt=\int\frac{2 \left(1-t^4-\sqrt{1-t^4}\right)}{t^2 \left(t^4-1\right)}\,dt$$ $$I=2 \left(\frac{1-\sqrt{1-t^4}}{t}+F\left(\left.\sin ^{-1}(t)\right|-1\right)-E\left(\left.\sin^{-1}(t)\right|-1\right) \right)$$

So, for the definite integral $$J=\int_0^{\frac \pi 2}\frac{\sqrt{\sin (x)}}{1+\cos (x)}\,dx=2\left( K(-1)-E(-1)+1\right)=2-\frac{2 \sqrt{2} \pi ^{3/2}}{\Gamma \left(\frac{1}{4}\right)^2}$$

Claude Leibovici
  • 289,558
1

$$\int_0^{\frac\pi 2}\left(\frac{\sqrt{\sin x}}{\cos x+1}+\sqrt{\sin x}\right)dx=\int_0^{\frac\pi 2}\frac{\cos x +2}{\cos x+1}\sqrt{\sin x}dx=\int_0^{\frac\pi 2}\frac{3(\sin\tfrac x2)^{\tfrac12}\cos\tfrac x2(\cos\tfrac x2)^\tfrac12+(\sin\tfrac x2)^\tfrac32(\cos x)^{-\tfrac12}\sin\tfrac x2}{\sqrt2(\cos^{\tfrac12}\tfrac x2)^2}dx =\frac{4}{\sqrt2}\frac{(\sin\tfrac x2)^{\tfrac32}}{(\cos\tfrac x2)^{\tfrac12}}\big\vert_0^{\tfrac\pi 2}=2$$ Hence, $$I=2-...$$

Bob Dobbs
  • 15,712
1

Thanks to Bob's insightful observation that

$$I + \int_0^\tfrac\pi2 \sqrt{\sin x} \, dx = \left[ \left(\sin x\right)^{3/2} \sec^2\frac x2 \right] \bigg|_0^\tfrac\pi2 = 2$$

we can connect the $x$- and $t$-integrals by substituting $\sin x=\cos^2t$,

$$\int_0^\tfrac\pi2 \sqrt{\sin x} \, dx = \int_0^\tfrac\pi2 \sqrt{\cos^2t} \frac{2\sin t\cos t}{\sqrt{1-\cos^4t}} \, dt = \int_0^\tfrac\pi2 \frac{2\cos^2t}{\sqrt{1+\cos^2t}}\,dt$$

user170231
  • 25,320