Integral $-\int_{0}^{a}{at^{3}\,{\rm d}t\over\,\sqrt{\,\left(a^{2} -t^{2}\right)\left(b^{2}t^{2} +a^{4}-a^{2}t^{2}\right)\,}\,}$

Question

This problem's taking me a lot longer than I like (probably doing things the hard way...). I have 9 different terms to integrate, some of which are messier than others. This one, though, is messier than most. $$-{\Large\int}_{\!0}^{a} \!\raise 0.8ex {at^{3}\,{\rm d}t \over \,\sqrt{\,\left(a^{2} -t^{2}\right)\left(b^{2}t^{2} +a^{4}-a^{2}t^{2}\right)\,}\,} $$

I tried a whole series of substitutions on this. First $t=a\sin\left(u\right)$, then $v=\cos\left(u\right)$ and split the resulting integral into two, only to find (after substitution number three) that the first half diverged (haven't tried the second half). In any case, I'm assuming there has to be a better way to attempt this. Any suggestions appreciated.

Answer 1

Using the substitution $u=a^2-t^2$ as suggested by André Nicolas, we get:

$\dfrac{\mathrm du}{\mathrm dt}=-2t$. So,

$\begin{align}&\int-\frac{at^3}{\sqrt{(a^2-t^2)(b^2t^2+a^4-a^2t^2)}}\mathrm dt\\ &=\frac{a}2\int\frac{t^2(-2t)}{\sqrt{(a^2-t^2)((b^2-a^2)t^2+a^4)}}\mathrm dt\\ &=\frac{a}2\int\frac{(a^2-u)}{\sqrt{u((b^2-a^2)(a^2-u)+a^4)}}\mathrm du\\ &=\frac{a}2\int\frac{(a^2-u)}{\sqrt{u(m-nu)}}\mathrm du \end{align}$

where $m=a^2b^2$ and $n=(b^2-a^2)$

$\begin{align} &=\frac{a^3}2\int\frac{1}{\sqrt{u(m-nu)}}\mathrm du - \frac{a}2\int\frac{\sqrt{u}}{\sqrt{m-nu}}\mathrm du\\ &=\frac{a^3}2I_1 - \frac{a}2I_2. \end{align}$

where I have substituted $I_1$ and $I_2$ for the first and second integral respectively.

For $I_1$, substitute $v=\sqrt{u}$, so that $\dfrac{\mathrm dv}{\mathrm du}=\dfrac1{2\sqrt{u}}$. Then,

$\displaystyle I_1=2\int \frac1{\sqrt{m-nv^2}}\mathrm dv =\frac2{\sqrt{m}}\int \frac1{\sqrt{1-{\left(\sqrt{\frac{n}{m}}v\right)}^2}}\mathrm dv =\dfrac{2}{\sqrt{m}}\sqrt{\dfrac{m}{n}}\arcsin \sqrt{\dfrac{n}{m}}v+C$.

Now for $I_2$,

$\begin{align} I_2=&\int\frac{\sqrt{u}}{\sqrt{m-nu}}\mathrm du =\frac1{\sqrt{m}}\sqrt{\frac{m}{n}}\int\frac{\sqrt{\frac{n}{m}u}}{\sqrt{1-\frac{n}{m}u}}\mathrm du\\ &=\frac1{\sqrt{m}}\sqrt{\frac{m}{n}}\frac{m}{n}\int\frac{\sqrt{x}}{\sqrt{1-x}}\mathrm dx \qquad\qquad{\left(x=\dfrac{n}{m}u\right)} \end{align}$

To solve $\displaystyle\int\frac{\sqrt{x}}{\sqrt{1-x}}\mathrm dx$, let $\theta=\arcsin \sqrt{x} (\implies x=\sin^2 \theta)$. We get, $\dfrac{\mathrm d\theta}{\mathrm dx}=\dfrac1{\sqrt{1-x}}\cdot\dfrac1{2\sqrt{x}}$.

So,

$\begin{align}&\int \sqrt{\frac{x}{1-x}} \mathrm dx =\int 2x \dfrac1{\sqrt{1-x}}\cdot\dfrac1{2\sqrt{x}}\mathrm dx\\ &=\int 2\sin^2\theta \;\mathrm d\theta =\theta-\dfrac{\sin2\theta}{2}+C\\ &=\arcsin \sqrt{x}-\dfrac12\sin(2\arcsin \sqrt{x})+C=\arcsin \sqrt{x}-\sin(\arcsin \sqrt{x})\cos (\arcsin \sqrt{x})+C\\ &=\arcsin \sqrt{x}-\sin(\arcsin \sqrt{x})\sqrt{1-\sin^2 (\arcsin \sqrt{x})}+C\\ &=\arcsin \sqrt{x}-\sqrt{x}\sqrt{1-x}+C. \end{align}$

Answer 2

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{\fermi\pars{a,b} \equiv -\int_{0}^{a}{at^{3}\,{\rm d}t \over \,\sqrt{\,\left(a^{2} - t^{2}\right)\left(b^{2}t^{2} +a^{4}- a^{2}t^{2}\right)\,}\,} \,,\quad \fermi\pars{0,b} = 0}$

When $a \not= 0$: \begin{align} \\[3mm] \fermi\pars{a,b} &= -\sgn\pars{a}\int_{0}^{\verts{a}}{\verts{a}t^{3}\,{\rm d}t \over \,\sqrt{\,\left(a^{2} - t^{2}\right)\left(b^{2}t^{2} +a^{4}- a^{2}t^{2}\right)\,}\,} \\[3mm]&= -a\verts{a}{\rm F}\pars{\mu}\quad\mbox{where}\quad{\rm F}\pars{\mu} \equiv\int_{0}^{1}{t^{3}\,{\rm d}t \over \,\sqrt{\,\left(1 - t^{2}\right)\left(\mu t^{2} + 1\right)\,}\,}\,,\quad \mu \equiv \pars{b \over a}^{2} - 1 \end{align}

\begin{align} {\rm F}\pars{\mu} &= \half\int_{0}^{1}{t\,{\rm d}t \over \root{\pars{1 - t}\pars{\mu t + 1}}} \end{align}

It's clear that $\color{#00f}{{\rm F}\pars{0} = 2/3}$. When $\color{#00f}{\mu \not= 0}$ we perform the change of variables $t = 1 - x^{2}$ $\quad\iff\quad$ $x = \pars{1 - t}^{1/2}$: \begin{align} {\rm F}\pars{\mu} &= \half\int_{1}^{0}{1 - x^{2} \over x\bracks{\mu\pars{1 - x^{2}} + 1}}\, \pars{-2x\,\dd x} = \int^{1}_{0}{1 - x^{2} \over \mu + 1 - \mu x^{2}}\,\dd x \\[3mm]&= {1 \over \mu}\int^{1}_{0}{\pars{\mu + 1- \mu x^{2}} - 1 \over \mu + 1 - \mu x^{2}}\,\dd x ={1 \over \mu} - {1 \over \mu^{2}}\int_{0}^{1}{\dd x \over \pars{1 + 1/\mu} - x^{2}} \end{align}

1. $\color{#00f}{\large \mu < 0}$: \begin{align} {\cal F}\pars{\mu} &=\color{#00f}{\large{1 \over \mu} + {1 \over \mu^{2}}\int_{0}^{1} {\dd x \over x^{2} + \pars{\root{-1 - 1/\mu}}^{2}}} \\[3mm]&={1 \over \mu} + {1 \over \mu^{2}}\,\root{-\mu \over \mu + 1} \arctan\pars{\root{-\mu \over \mu + 1}} \end{align}
2. $\color{#00f}{\large \mu > 0}$: \begin{align} {\cal F}\pars{\mu} &={1 \over \mu} + {1 \over \mu^{2}}\int_{0}^{1} {\dd x \over x^{2} - \pars{\root{1 + 1/\mu}}^{2}} \\[3mm]&={1 \over \mu} + {1 \over \mu^{2}} \int_{0}^{1}\pars{{1 \over x - \root{1 + 1/\mu}} - {1 \over x + \root{1 + 1/\mu}}} {1 \over 2\root{1 + 1/\mu}} \\[3mm]&=\color{#00f}{\large{1 \over \mu} + {1 \over 2\mu^{2}}\, \root{\mu \over 1 + \mu} \ln\pars{\verts{\root{\mu} - \root{1 + \mu} \over \root{\mu} + \root{1 + \mu}}}} \end{align}
3. $\color{#00f}{\large \mu = 0}$: $${\cal F}\pars{\mu} = \color{#00f}{\large{\cal F}\pars{0} = {2 \over 3}} $$