G'day, mate. I wonder... Is it possible to obtain the value of this integral? $$\int_0^\pi \sqrt{1+b^2\sin^2(x)}\,dx,\text{where b is positive real number.} $$ And I turned to Mathematica's aid, and it gave me the result: $$2E(-b^2) $$ where E is the complete elliptic integral of the second kind;(why?? but its upper limit is π instead of π/2.) It didn't give me a step-by-step solution, so I still don't understand how it solved it. Plus, I doubt if this result is incorrect, since even rewriting the integral as:$$\int_0^\pi \sqrt{1-(-b^2)\sin^2(x)}\,dx$$it still doesn't match the form of the complete elliptic integral of the second kind:$$\int_0^{\pi/2}\sqrt{1-k^2\sin^2(x)}\,dx$$The most confusing part is that$$k^2=-b^2$$ it's impossible since b is a positive real number! I'm so confused, hope someone can help me out. Thanks a lot!
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Hint, what is $\sin(\pi-x)$ and why is this the double of the integral from $0$ to $\frac{\pi}2$? – Raymond Manzoni Dec 02 '22 at 10:01
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Have you tried using the half-angle formula? $\sin^2(x)=\frac{1-\cos(2x)}{2}$. I don't know if it answers your question, it's just an idea :) – Martingalo Dec 02 '22 at 10:02
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btw for mathematica $E\left(\varphi ,|,k^2\right) = E(\sin\varphi;k) = \int_0^\varphi \sqrt{1-k^2 \sin^2\theta}, d\theta$ so that replacing $k^2$ by $-k^2\cdots$ – Raymond Manzoni Dec 02 '22 at 10:07
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@Martingalo thank you but if go this way,$$\sqrt{(1+\frac{b^2}{2})-\frac{b^2}{2}\cos(2x)},dx$$, we still can’t open the square root. – 106207436 Dec 03 '22 at 09:51
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@RaymondManzoni hi, hmm I’m not familiar with the elliptic integrals, so I guess you mean $$2E(-b^2)$$ is correct, right? – 106207436 Dec 03 '22 at 10:03
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Yes! My first tip about $\sin(\pi-x)$ should help you to rewrite your integral as the integral over $(\frac{\pi}2,\pi)$ added to the integral over $(0,\frac{\pi}2)$ (prove that they are equal). By definition $E\left(\varphi ,|,-b^2\right) = \int_0^\varphi \sqrt{1+b^2 \sin^2\theta}, d\theta$ so that mma is right! To obtain $k^2=-b^2$ you need $b=\pm i k$ so that you are in fact replacing $k$ by (say) $i b$. Hoping this clarified things, – Raymond Manzoni Dec 03 '22 at 10:27
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1@RaymondManzoni hi, I came up with a method and finally solved it! But my answer is $$2\sqrt{1+b^2}E(\frac{b}{\sqrt{1+b^2}})$$ – 106207436 Dec 03 '22 at 14:14
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In fact $\displaystyle E(i b)=2\sqrt{1+b^2}E\left(\frac{b}{\sqrt{1+b^2}}\right)$ if you use the definition $E(k):=\int_0^\varphi \sqrt{1-k^2 \sin^2\theta}, d\theta$ (and not mathematica's definition writing this as $E(k^2)$ i.e. squaring the parameter of $E$) – Raymond Manzoni Dec 03 '22 at 14:38
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(mma and alpha use the parameter $m=k^2$ convention while more usual definitions use the modulus $k$. This is detailed in JM's answer here ) – Raymond Manzoni Dec 03 '22 at 14:47
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Mma's convention for the previous functional identity would thus be $\displaystyle E[ -b^2]=\sqrt{1+b^2}E\left[\frac{b^2}{1+b^2}\right];$ but this is simply the second identity here for $z:=-b^2$. (sorry my initial functional identity should have been $\displaystyle E(i b)=\sqrt{1+b^2}E\left(\frac{b}{\sqrt{1+b^2}}\right)$ without the $2$). – Raymond Manzoni Dec 03 '22 at 14:56
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@RaymondManzoni thanks for your conscientious discussion! – 106207436 Dec 05 '22 at 11:15
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Glad it helped @MilneAndo ! – Raymond Manzoni Dec 05 '22 at 12:52
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Let$$\chi=\int^\pi_0 \sqrt{1+b^2\big[1-\cos^2(x)\big]}{\rm d}x= \int^\pi_0 \sqrt{1+b^2}\sqrt{1-\frac{b^2}{1+b^2}\cos^2(x)}\,{\rm d}x$$ Here let$$k^2=\frac{b^2}{1+b^2}$$and separate the integral into two parts, so $$\chi=\sqrt{1+b^2}\bigg[\int^\frac{\pi}{2}_0 \sqrt{1-k^2\cos^2(x)\,}{\rm d}x+ \int^\pi_\frac{\pi}{2} \sqrt{1-k^2\cos^2(x)\,}{\rm d}x\bigg]$$ For the former one, let $$\cos(x)\big|^\frac{\pi}{2}_0 =\sin\big(\frac{\pi}{2}-x\big)=\sin(z')\big|^0_\frac{\pi}{2}$$ For the latter one, let $$\cos(x)\big|^\pi_\frac{\pi}{2} =-\sin\big(x-\frac{\pi}{2}\big)=-\sin(z)\big|^\frac{\pi}{2}_0$$ After some efforts and obtain $$\chi=\sqrt{1+b^2}\big[E(k)+E(k)\big]$$
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