TITLE: Is there a nice way to find this integral $\int_0^1\frac{ \arcsin x}{x} \mathrm{d}x$?
QUESTION [8 upvotes]: $$\int_0^1\frac{ \arcsin x}{x}\,\mathrm dx$$
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.

REPLY [3 votes]: 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
$$\int_{0}^{\frac{\pi }{2}}\frac{x}{\tan x}dx=\frac{\pi }{2}\ln(2).$$