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I'm hopelessly stuck with the integral

$$\int_{-1}^1 \frac{1}{(x-2)(x+2)\sqrt{1-x^2}} dx$$

I'm guessing the way to go is complex contour integration, but I have no idea.

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    dogbone contour – 2017-02-10

3 Answers 3

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If you set $x=\sin\arctan t=\frac{t}{\sqrt{t^2+1}}$ you get an elementary integral: $$ \int_{-\infty}^{+\infty}\frac{dt}{t^2-4(1+t^2)} = \color{red}{-\frac{\pi}{2\sqrt{3}}}.$$

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    Where do you get the infinite boundaries from? – 2017-02-09
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    @con-f-use: from the substitution I used. If $x$ originally ranges from $-1$ to $1$ and $x=\sin\arctan(t)$, then $t$ ranges from $-\infty$ to $+\infty$. No magic, just the usual substitutions machinery. – 2017-02-09
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    Thanks for taking the time to answer. How do you reconcile that with $x=\frac{t}{t^2+1}$? – 2017-02-09
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    @con-f-use: $\sin\arctan t=\frac{t}{\sqrt{t^2+1}}$ is a basic trigonometric identity. Draw a right triangle with two legs having length $1$ and $t$ and make some straightforward consideration about its angles. – 2017-02-09
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    I'm just saying, if you computed the boundaries via $x=\frac{t}{\sqrt{t^2+1}}$, you'd get a different answer. But I'm obviously being stupid, so again thanks for your patience. – 2017-02-09
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    @con-f-use $-1=\sin\arctan t$ so $-\frac{\pi}{2}=\arctan t$ then $-\infty=t$. – 2017-02-09
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    @con-f-use: why a different answer? The only point for which $\frac{t}{\sqrt{t^2+1}}=1$ is $t=+\infty$ (informally speaking, since $+\infty$ is not a number) and the only point for which $\frac{t}{\sqrt{t^2+1}}=-1$ is $t=-\infty$ (again, informally speaking). – 2017-02-09
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Let $x=\sin t$ $$\int_{-1}^1 \frac{1}{(x-2)(x+2)\sqrt{1-x^2}} dx=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{1}{\sin^2t-4}dt$$ and then $\sin t=\dfrac{2\tan\frac{t}{2}}{1+\tan\frac{t}{2}}$.

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    Not sure how the tan thing. Do you mean a second substituion? – 2017-02-09
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    Note, with complex contour integration it has a simple way, but in my way you need another substitution $\tan=u$ and maybe another one after it for separate fractions. – 2017-02-09
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    Are you actually sure it works like that? Thanks for you help in any case. – 2017-02-09
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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} &\int_{-1}^{1}{\dd x \over \pars{x - 2}\pars{x + 2}\root{1 - x^{2}}} = 2\int_{0}^{1}{\dd x \over \pars{x^{2} - 4}\root{1 - x^{2}}} \\[5mm] \stackrel{x\ \mapsto\ 1/x}{=}\,\,\, & \int_{1}^{\infty}{2x\,\dd x \over \pars{1 - 4x^{2}}\root{x^{2} - 1}} \,\,\,\stackrel{x^{2}\ \mapsto\ x}{=}\,\,\, \int_{1}^{\infty}{\dd x \over \pars{1 - 4x}\root{x - 1}} \\[5mm] \stackrel{x\ =\ t^{2} + 1}{=}\,\,\, & 2\int_{0}^{\infty}{\dd t \over -3 - 4t^{2}} = \bbx{\ds{-\,{\root{3} \over 6}\,\pi}} \end{align}

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    Good job. Simple and useful for @con-f-use. – 2017-02-09
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    @MyGlasses Thanks. – 2017-02-11