Compute $\int_0^1 \frac{\arcsin(x)}{x}dx$

$$\int_0^1 \frac{\arcsin(x)}{x}dx$$

This is a proposed for a Calculus II exam, and I have absolutely no idea how to solve it. Tried using Frullani or Lobachevsky integrals, or beta and gamma functions, but I can't even find a way to start it. Wolfram Alpha gives a kilometric solution, but I know that cannot be the only answer. Any help appreciated!

• Maple gives an answer of $\frac{\pi}{2} \ln (2)$. – i. m. soloveichik Jun 7 '14 at 14:11
• Integrate by part first. If you don't know how to compute the remaining integral, look at answers of this question. – achille hui Jun 7 '14 at 14:15

Let $y=\arcsin x\;\Rightarrow\;\sin y =x\;\Rightarrow\;\cos y\ dy=dx$, then $$\int_0^1 \frac{\arcsin(x)}{x}dx=\int_0^{\Large\frac\pi2}y\cot y\ dy.$$ Now use IBP by taking $u=y$ and $dv=\cot y\ dy\;\Rightarrow\;v=\ln(\sin x)$, then \begin{align} \int_0^{\Large\frac\pi2}y\ \cot y\ dy&=\left.y\ln(\sin y)\right|_0^{\Large\frac\pi2}-\int_0^{\Large\frac\pi2}\ln(\sin y)\ dy\\ &=-\int_0^{\Large\frac\pi2}\ln(\sin y)\ dy. \end{align} The last integral can be evaluated by using property $$\int_a^b f(x)\ dx=\int_a^b f(a+b-x)\ dx.$$ We obtain $$\int_0^{\Large\frac\pi2}\ln(\sin y)\ dy=-\frac\pi2\ln2,$$ where $$\int_0^{\Large\frac\pi2}\ln(\sin y)\ dy=\int_0^{\Large\frac\pi2}\ln(\cos y)\ dy\quad\Rightarrow\quad\text{by symmetry}.$$ Thus $$\int_0^1 \frac{\arcsin(x)}{x}dx=\large\color{blue}{\frac\pi2\ln2}.$$