Is there a nice way to find this integral $\int_0^1\frac{ \arcsin x}{x} \mathrm{d}x$? $$\int_0^1\frac{ \arcsin x}{x}\,\mathrm dx$$
I was looking in my calculus text by chance when I saw this example , the solution  is written also but it uses very tricky methods for me ! I wonder If there is a nice way to find this integral.
The idea of the solution in the text is in brief , Assume $y=\sin(x)$ and use definition of improper integral and some properties  of definite integral to get $ -\lim_{\varepsilon \rightarrow0^+} \int_\epsilon^{\frac{\pi}{2}} \ln(\cos(y-\varepsilon))\,dy$
then it uses that fact that $a= (a+a) \times \frac{1}{2}$ where $a$ here is the integral , and then there is a step which I can't understand till the moment( but understand most of the rest of the steps ) . and it keeps going to use more and more tricks to get the final result , $\frac{\pi}{2} \ln2$.
Now, I try to understand this method , If I couldn't I will ask for help, but in the moment, I wonder if there is a good way to find this integral. 
 A: $$I = \int_0^1 \frac{\arcsin x}{x}\,dx = \int_{0}^{\pi/2}\frac{x\,dx}{\sin x}\cos x = x\log\sin x \bigg|_0^{\pi/2}-\int_0^{\pi/2} \log\sin x\,dx$$
This last integral is well known to equal $-\frac{\pi}{2}\log 2$.  Thus
$$I = \frac{\pi}{2}\log 2$$
A: Using integration by parts with $u=\arcsin(x)$ yields

$$ \int_0^1 \frac{\arcsin x}{x}\,dx = -\int_0^1 \frac{\ln(x)}{\sqrt{1-x^2}}\,dx=I .$$

Now, consider the integral

$$ F = \int_0^1 \frac{x^\alpha}{\sqrt{1-x^2}}\,dx = \frac{\sqrt{\pi}}{2}{\frac {\Gamma  \left( \frac{\alpha}{2}+\frac{1}{2} \right) }{
\Gamma  \left( \frac{\alpha}{2}+1 \right) }},$$

which can be evaluated using the $\beta$ function (subs $x^2=t$ in F). Now, $I$ follows from $F$ as

$$ I = \lim_{\alpha\to 0}\frac{dF}{d\alpha}= \frac{\pi \ln(2)}{2}.$$

A: Let $$ y=sin^{-1}x\ \ \Rightarrow x= \sin y,\ \ dx= \cos y dy,$$
then
$$
I=\int_{0}^{1}\frac{sin^{-1}x}{x}dx=\int_{0}^{\frac{\pi }{2}}\frac{y}{siny}.cosy=\int_{0}^{\frac{\pi }{2}}\frac{y}{tany}dy.$$
Let
$$I(a)=\int_{0}^{\frac{\pi }{2}}\frac{\arctan(a \tan y)}{\tan y}dy \Rightarrow I'(a)=\int_{0}^{\frac{\pi }{2}}\frac{dy}{1+(a \tan(y))^2}.$$
One can easily prove that $$\int_{0}^{\frac{\pi }{2}}\frac{1}{1+(a \tan(y))^2}dy=\frac{\pi }{2(1+a)}.$$
Therefore 
$$I'(a)=\frac{\pi }{2(1+a)}\ \ \Rightarrow I(a)=\frac{\pi }{2}\ln(1+a)+c$$
and therefore 
$$I(a)=\int_{0}^{\frac{\pi }{2}}\frac{\arctan(a \tan x)}{\tan x}dx=\frac{\pi }{2}\ln(1+a)+c.$$
Substitute $a=0$ to find $c=0,$ therefore 
$$I(a)=\frac{\pi }{2}\ln(1+a).$$
Now putting $a=1,$ we get
$$\int_{0}^{\frac{\pi }{2}}\frac{x}{\tan x}dx=\frac{\pi }{2}\ln(2).$$
A: Let $$I=\int_{0}^{1} \frac{\arcsin(x)}{x} dx.$$ Put $x=\sin(\theta), dx=\cos(\theta) d\theta$ so that $$I=\int_{0}^{\frac{\pi}{2}}\frac{\theta}{\tan(\theta)} d\theta.$$ We will show this representation of $I$ is indeed equal to a double integral that we can easily evaluate in two ways.  
Consider $$J=\int_{0}^{1}\int_{0}^{\infty}\frac{1}{(1+x^2y^2)(1+y^2)} dydx.$$ Using partial fractions, you can integrate with respect to $y$ first as such:
$$\frac{1}{(1+x^2y^2)(1+y^2)}=\frac{1}{1-x^2} \left(\frac{-x^2}{1+x^2y^2}+\frac{1}{1+y^2}\right).$$ so $$J=\int_{0}^{1} \frac{1}{1-x^2} \left( -\frac{\pi}{2}x +\frac{\pi}{2} \right)dx=\int_{0}^{1} \frac{\pi}{2}\left(\frac{1}{1+x}\right) dx=\frac{\pi \ln(2)}{2}.$$ On the other hand, let us try to integrate $J$ the other away around as such:
$$J=\int_{0}^{\infty}\int_{0}^{1}\frac{1}{(1+x^2y^2)(1+y^2)} dxdy.$$
$$J=\int_{0}^{\infty} \frac{\tan^{-1}(y)}{y(y^2+1)}dy.$$ Now let $y=\tan(\theta),dy=\sec^{2}(\theta) d\theta$ so that we get:
$$J=\int_{0}^{\frac{\pi}{2}}\frac{\theta}{\tan(\theta)} d\theta,$$ so we get that $$J=I=\frac{\pi \ln(2)}{2}.$$
