Calculate $\int_{0}^{1}\frac{\arctan(x)}{x\sqrt{1-x^2}}dx$ I am preparing for a calculus exam and I was asked to calculate $$\int_{0}^{1}\frac{\arctan(x)}{x\sqrt{1-x^2}}dx$$
Using the hint that $$\frac{\arctan(x)}{x}=\int_{0}^{1}\frac{dy}{1+x^2y^2}$$
I ran into some trouble and would appreciate help.
What I did:
I used the hint, $$\int_{0}^{1}\frac{\arctan(x)}{x\sqrt{1-x^2}}dx=\int_{0}^{1}\int_{0}^{1}\frac{1}{(1+x^2y^2)\sqrt{1-x^2}}dydx$$
Since $x,y$ move between $0,1$ I thought maybe it is best to used the transform $x=r\cos\theta$ , $y=r\sin\theta$. $r \in [0,1]$, $\theta \in [0, \frac{\pi}{2}]$
But I didn't get anything meaningful, I didn't end up an something that is easy / possible to integrate. And I honestly can't think of a way to integrate this as it is.
 A: May I show another way?. It does not use the given hint though.
As Norbert done, make the sub $\displaystyle x=\sin(t), \;\ dx=\cos(t)dt$
$$\int_{0}^{\frac{\pi}{2}}\frac{\tan^{-1}(\sin(t))}{\sin(t)}dt$$
Now, use the series for arctan: $\displaystyle \tan^{-1}(x)=\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{2n+1}}{2n+1}$
only, $x=\sin(t)$ and we get:
$$\int_{0}^{\frac{\pi}{2}}\sum_{n=0}^{\infty}\frac{\sin^{2n}(t)(-1)^{n}}{2n+1}dt$$
But, using the famous and known result: $\displaystyle \int_{0}^{\frac{\pi}{2}}\sin^{2n}(t)dt=\frac{\pi (2n)!}{2^{2n+1}(n!)^{2}}$, one obtains the series:
$$\frac{\pi}{2}\sum_{n=0}^{\infty}\frac{(-1)^{n} (2n)!}{(2n+1)2^{2n}(n!)^{2}}$$
Now, notice what that series is?.  It is the Taylor series for $\displaystyle \sinh^{-1}(1)$
which is $$\sinh^{-1}(x)=\sum_{n=0}^{\infty}\frac{(-1)^{n}(2n)!x^{2n+1}}{2^{2n}(n!)^{2}(2n+1)}$$
so, we have $$\frac{\pi}{2}\sinh^{-1}(1)$$
A: $\newcommand{\+}{^{\dagger}}
 \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
 \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
 \newcommand{\dd}{{\rm d}}
 \newcommand{\down}{\downarrow}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\fermi}{\,{\rm f}}
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\isdiv}{\,\left.\right\vert\,}
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\sech}{\,{\rm sech}}
 \newcommand{\sgn}{\,{\rm sgn}}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}
 \newcommand{\wt}[1]{\widetilde{#1}}$
$\ds{\int_{0}^{1}{\arctan\pars{x} \over x\root{1 - x^{2}}}\,\dd x:\ {\large ?}}$

Let's consider
  $\ds{\fermi\pars{\mu} \equiv
\int_{0}^{1}{\arctan\pars{\mu x} \over x\root{1 - x^{2}}}\,\dd x\,,\quad\fermi\pars{0} = 0\,,\qquad\fermi\pars{1}={\large ?}}$:

\begin{align}
\fermi'\pars{\mu}&=
\int_{0}^{1}{\dd x \over \pars{\mu^{2}x^{2} + 1}\root{1 - x^{2}}}
=\int_{0}^{\pi/2}{\dd\theta \over \mu^{2}\cos^{2}\pars{\theta} + 1}
=\int_{0}^{\pi/2}{\sec^{2}\pars{\theta}\,\dd\theta\over \mu^{2} + 1 + \tan^{2}\pars{\theta}}
\\[3mm]&=\int_{0}^{\infty}{\dd t \over t^{2} + 1 + \mu^{2}}
={1 \over \root{1 + \mu^{2}}}\int_{0}^{\infty}{\dd t \over t^{2} + 1}
={\pi \over 2}\,{1 \over \root{1 + \mu^{2}}}
\end{align}

\begin{align}
\fermi\pars{1}&={\pi \over 2}\
\overbrace{\int_{0}^{1}{\dd\mu \over \root{1 + \mu^{2}}}}
^{\ds{\mbox{Set}\ \mu \equiv \sinh\pars{\theta}}}
={\pi \over 2}\,{\rm arcsinh}\pars{1}
\end{align}

$$\color{#66f}{\large%
\int_{0}^{1}{\arctan\pars{x} \over x\root{1 - x^{2}}}\,\dd x
={\pi \over 2}\,{\rm arcsinh}\pars{1}} \approx 1.3845
$$
A: $$
\begin{align}
\int_{0}^{1}\int_{0}^{1}\frac{1}{(1+x^2y^2)\sqrt{1-x^2}}dydx
&=\int_{0}^{1}\int_{0}^{1}\frac{1}{(1+x^2y^2)\sqrt{1-x^2}}dxdy\\
&=\int_{0}^{1}\int_{0}^{\pi/2}\frac{1}{1+y^2 \sin^2 t}dtdy\\
&=\int_{0}^{1}\int_{0}^{\pi/2}\frac{1}{\cos^2 t+ (1+y^2)\sin^2 t}dtdy\\
&=\int_{0}^{1}\int_{0}^{\pi/2}\frac{1}{1 + (1+y^2)\tan^2 t}\frac{1}{\cos^2 t}dtdy\\
&=\int_{0}^{1}\int_{0}^{\infty}\frac{1}{1 + (1+y^2)s^2}dsdy\\
&=\int_{0}^{1}\frac{1}{1+y^2}\int_{0}^{\infty}\frac{1}{\left((1+y^2)^{-1/2}\right)^2+s^2}dsdy\\
&=\int_{0}^{1}\frac{1}{1+y^2}\sqrt{1+y^2}\arctan \left(s\sqrt{1+y^2}\right)\Biggl|_0^\infty dy\\
&=\int_{0}^{1}\frac{\pi}{2\sqrt{1+y^2}}dy\\
&=\frac{\pi}{2}\operatorname{arcsinh} y\Biggl|_0^1\\
&=\frac{\pi}{2}\operatorname{arcsinh} 1
\end{align}
$$
A: The following is a different approach that doesn't use that hint.
$$ \begin{align} \int_{0}^{1} \frac{\arctan (x)}{x\sqrt{1-x^{2}}} \ dx &= \int_{0}^{1} \frac{1}{x \sqrt{1-x^{2}}} \sum_{n=0}^{\infty}  \frac{(-1)^{n}}{2n+1} x^{2n+1} \ dx \tag{1}\\ &= \sum_{n=0}^{\infty} \frac{(-1)^{n}}{2n+1} \int_{0}^{1} \frac{x^{2n}}{\sqrt{1-x^{2}}} \ dx  \\ &= \frac{1}{2} \sum_{n=0}^{\infty} \frac{(-1)^{n}}{2n+1} \int_{0}^{1} u^{(n+1/2)-1} (1-u)^{1/2-1} \ du \tag{2} \\ &= \frac{1}{2} \sum_{n=0}^{\infty} \frac{(-1)^{n}}{2n+1} B\left(n+\frac{1}{2}, \frac{1}{2} \right) \tag{3}\\ &=\frac{1}{2} \sum_{n=0}^{\infty} \frac{(-1)^{n}}{2n+1} \frac{\Gamma (n+\frac{1}{2}) \Gamma(\frac{1}{2})}{\Gamma (n+1)} \tag{4} \\ &= \frac{1}{2}\sum_{n=0}^{\infty} \frac{(-1)^{n}}{2n+1} \frac{\Gamma (\frac{1}{2})}{\Gamma (n+1)} \frac{\Gamma(2n) \Gamma (\frac{1}{2})}{\Gamma (n) 2^{2n-1}} \frac{2n}{2n}  \tag{5} \\ &= \frac{\pi}{2} \sum_{n=0}^{\infty} \frac{(-1)^{n} \Gamma(2n+1)}{2^{2n} \ \Gamma^{2}(n+1)} \frac{1}{2n+1} \tag{6} \\ &= \frac{\pi}{2} \text{arcsinh}(1) \tag{7}\end{align}$$
$ $
(1) Maclaurin expansion of arctan(x)
(2) let $u = x^{2}$
(3) integral representation of the beta function
(4) defintion of the beta function in terms of the gamma function
(5) gamma function duplication formula
(6) $\Gamma(z+1) = z \Gamma(z)$ and $\Gamma(\frac{1}{2}) = \sqrt{\pi}$
(7) Maclaurin expansion of arcsinh(x)
