Find the integral for equation

Question

What is $$\int_{-x_t}^0\sqrt{\frac1{1+(x/h)^2}-b}\,dx$$ for $h>0,0<b<1$?

Mathematica gave $$ih\sqrt{1-b}\cdot E\left(i\sinh^{-1}\log\frac b2,\frac b{b-1}\right)\qquad2\sqrt{-1+(1/b)}+\log b>0$$ but I want the answer without using it.

Answer 1

As @Peter commented, using $x=h\sinh(t)$ the integral becomes $$I=h \int \sqrt{1-b \cosh ^2(t)}\,dt$$ Now $t=i \,u$ gives $$I=i\,h \int \sqrt{1-b \cos ^2(u)}\,du=i\,h \sqrt{1-b}\,\, E\left(u\left|-\frac{b}{1-b}\right.\right)$$

Comment

I wonder if there could be a mistake in what you report in the post. Repeating the calculations, what I obtained is

$$i\, \sqrt{1-b}\, h\,\, E\left(i \sinh ^{-1}\left(\frac{x_t}{h}\right)|-\frac{b}{1-b}\right)$$

Answer 2

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Given $\left(a,b,x_{1}\right)\in\mathbb{R}^{3}$ such that $0<a\land0<b<1$ and $-a\sqrt{\frac{1-b}{b}}\le x_{1}\le a\sqrt{\frac{1-b}{b}}$, let $\mathcal{I}{\left(a,b,x_{1}\right)}$ denote the value of the definite integral

$$\mathcal{I}{\left(a,b,x_{1}\right)}:=\int_{0}^{x_{1}}\mathrm{d}x\,\sqrt{(1+(x/a)^{2})^{-1}-b}.$$

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Suppose $a\in\mathbb{R}_{>0}\land b\in\left(0,1\right)\land x_{1}\in\left[-a\sqrt{\frac{1-b}{b}},a\sqrt{\frac{1-b}{b}}\right]$.

Setting $z:=\frac{x_{1}}{a\sqrt{\frac{1-b}{b}}}\land p:=\sqrt{\frac{b}{1-b}}$, we have $-1\le z\le1\land0<p\land b=\frac{p^{2}}{p^{2}+1}$, and then

$$\begin{align} \mathcal{I}{\left(a,b,x_{1}\right)} &=\int_{0}^{x_{1}}\mathrm{d}x\,\sqrt{(1+(x/a)^{2})^{-1}-b}\\ &=a\sqrt{\frac{1-b}{b}}\int_{0}^{\frac{x_{1}}{a\sqrt{\frac{1-b}{b}}}}\mathrm{d}y\,\sqrt{\left[1+\left(\frac{1-b}{b}\right)y^{2}\right]^{-1}-b};~~~\small{\left[x=a\sqrt{\frac{1-b}{b}}\,y\right]}\\ &=a\sqrt{\frac{1-b}{b}}\int_{0}^{z}\mathrm{d}y\,\sqrt{\left[1+\left(\frac{1-b}{b}\right)y^{2}\right]^{-1}-b}\\ &=a\sqrt{1-b}\int_{0}^{z}\mathrm{d}y\,\sqrt{\frac{1-y^{2}}{\left(\frac{b}{1-b}\right)+y^{2}}}\\ &=\frac{a}{\sqrt{1+p^{2}}}\int_{0}^{z}\mathrm{d}y\,\sqrt{\frac{1-y^{2}}{p^{2}+y^{2}}}\\ &=\frac{a\operatorname{sgn}{\left(z\right)}}{\sqrt{1+p^{2}}}\int_{0}^{|z|}\mathrm{d}y\,\sqrt{\frac{1-y^{2}}{p^{2}+y^{2}}}\\ &=\frac{a\operatorname{sgn}{\left(z\right)}}{\sqrt{1+p^{2}}}\int_{\sqrt{1-z^{2}}}^{1}\mathrm{d}t\,\frac{t^{2}}{\sqrt{1-t^{2}}}\sqrt{\frac{1}{p^{2}+1-t^{2}}};~~~\small{\left[y=\sqrt{1-t^{2}}\right]}\\ &=aq\operatorname{sgn}{\left(z\right)}\int_{\sqrt{1-z^{2}}}^{1}\mathrm{d}t\,\frac{t^{2}}{\sqrt{1-t^{2}}}\sqrt{\frac{1}{q^{-2}-t^{2}}};~~~\small{\left[\frac{1}{\sqrt{1+p^{2}}}=:q\in(0,1)\right]}\\ &=a\operatorname{sgn}{\left(z\right)}\int_{\sqrt{1-z^{2}}}^{1}\mathrm{d}t\,\frac{q^{2}t^{2}}{\sqrt{1-t^{2}}\sqrt{1-q^{2}t^{2}}}\\ &=a\operatorname{sgn}{\left(z\right)}\int_{\sqrt{1-z^{2}}}^{1}\mathrm{d}t\,\left[\frac{1}{\sqrt{1-t^{2}}\sqrt{1-q^{2}t^{2}}}-\frac{\sqrt{1-q^{2}t^{2}}}{\sqrt{1-t^{2}}}\right]\\ &=a\operatorname{sgn}{\left(z\right)}\int_{0}^{1}\mathrm{d}t\,\left[\frac{1}{\sqrt{1-t^{2}}\sqrt{1-q^{2}t^{2}}}-\frac{\sqrt{1-q^{2}t^{2}}}{\sqrt{1-t^{2}}}\right]\\ &~~~~~-a\operatorname{sgn}{\left(z\right)}\int_{0}^{\sqrt{1-z^{2}}}\mathrm{d}t\,\left[\frac{1}{\sqrt{1-t^{2}}\sqrt{1-q^{2}t^{2}}}-\frac{\sqrt{1-q^{2}t^{2}}}{\sqrt{1-t^{2}}}\right]\\ &=a\operatorname{sgn}{\left(z\right)}\left[F{\left(\frac{\pi}{2},q\right)}-E{\left(\frac{\pi}{2},q\right)}\right]\\ &~~~~~-a\operatorname{sgn}{\left(z\right)}\left[F{\left(\arcsin{\left(\sqrt{1-z^{2}}\right)},q\right)}-E{\left(\arcsin{\left(\sqrt{1-z^{2}}\right)},q\right)}\right]\\ &=a\operatorname{sgn}{\left(z\right)}\left[K{\left(q\right)}-E{\left(q\right)}-F{\left(\arcsin{\left(\sqrt{1-z^{2}}\right)},q\right)}+E{\left(\arcsin{\left(\sqrt{1-z^{2}}\right)},q\right)}\right],\\ \end{align}$$

where

$$F{\left(\varphi,q\right)}:=\int_{0}^{\sin{(\varphi)}}\mathrm{d}t\,\frac{1}{\sqrt{1-t^{2}}\sqrt{1-q^{2}t^{2}}},$$

$$E{\left(\varphi,q\right)}:=\int_{0}^{\sin{(\varphi)}}\mathrm{d}t\,\frac{\sqrt{1-q^{2}t^{2}}}{\sqrt{1-t^{2}}},$$

and

$$K{\left(q\right)}:=F{\left(\frac{\pi}{2},q\right)},$$

$$E{\left(q\right)}:=E{\left(\frac{\pi}{2},q\right)}.$$

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