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Evaluate the integral $$J(d,x)=\int_0^x\frac{1}{\sqrt{d^2-\sin(r)-\cos(r)}}dr;\space d^2>\sqrt2$$
I am unsure where to begin with this. Any help appreciated.

mathreadler
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Kirin171
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    Mathematica gives a solution in terms of the Appell hypergeometric function (effectively a series solution). Doubtful that there's a "nice" closed form solution. – erfink May 10 '17 at 19:33
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    Do you mind telling about the context where this integral appeared? As it is written now, it looks like a random integral that is just made to be difficult. It seems that a primitive function is given by $$-\frac{2 F\left(\frac{5 \pi }{8}-\frac{r}{2}|\frac{2 \sqrt{2}}{d^2+\sqrt{2}}\right)}{\sqrt{d^2+\sqrt{2}}}$$ where $F$ is an Elliptic integral. – mickep May 11 '17 at 06:57
  • An old textbook, most of the problems seem to be as described. Nevertheless I'm having fun trying to figure them out. – Kirin171 May 11 '17 at 17:50

1 Answers1

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$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \left.\vphantom{\Large A}\mrm{J}\pars{d,x}\right\vert_{\ d^{2}\ >\ \root2} & \equiv \int_{0}^{x}{\dd r \over \root{d^{2} - \sin\pars{r} - \cos\pars{r}}} = \int_{0}^{x}{\dd r \over \root{d^{2} - \root{2}\cos\pars{r - \pi/4}}} \\[5mm] & = \int_{-\pi/4}^{x - \pi/4}{\dd r \over \root{d^{2} - \root{2}\cos\pars{r}}} = \int_{-\pi/4}^{x - \pi/4}{\dd r \over \root{d^{2} - \root{2}\bracks{1 - 2\sin^{2}\pars{r/2}}}} \\[5mm] & = {2 \over \root{d^{2} - \root{2}}}\int_{-\pi/8}^{x/2 - \pi/8} {\dd r \over \root{1 -\bracks{\root{2\root{2}/\pars{d^{2} - \root{2}2}}\,\ic}^{2} \sin^{2}\pars{r}}} \\[1cm] & = {2 \over \root{d^{2} - \root{2}}}\int_{-\pi/8}^{x/2 - \pi/8} {\dd r \over \root{1 -\bracks{\root{2\root{2}/\pars{d^{2} - \root{2}}}\,\ic}^{2} \sin^{2}\pars{r}}} \\ & + {2 \over \root{d^{2} - \root{2}}}\int_{0}^{\pi/8} {\dd r \over \root{1 -\bracks{\root{2\root{2}/\pars{d^{2} - \root{2}}}\,\ic}^{2} \sin^{2}\pars{r}}} \\[1cm] & = {2 \over \root{d^{2} - \root{2}}}\bracks{% \mrm{F}\pars{{x \over 2} - {\pi \over 8},\root{2\root{2} \over d^{2} - {2}}\,\ic} + \mrm{F}\pars{{\pi \over 8},\root{2\root{2} \over d^{2} - \root{2}}\,\ic}} \end{align}

$\ds{\mrm{F}}$ is a Legendre Integral. The above $\ds{\,\mrm{F}}$ arguments satisfy some conditions which are discussed in the above mentioned link.

Felix Marin
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