the integral similar to elliptic integral with upper limit is π instead of π/2

Question

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!

Answer 1

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 $$\chi=\sqrt{1+b^2}\big[E(k)+E(k)\big]$$