Calculate integrals $\int_0^1 {\frac{{\arcsin x}}{x}dx} $ This is my problem,I tried to use change of variable, but no result so far. Can anyone help me?
$$\int_0^1 {\frac{{\arcsin x}}{x}dx} 
$$
 A: Replace $\arcsin x$ by $\theta$ Then the integral becomes $$I=\int_{0}^{\pi/2} \frac{\theta}{\sin \theta}\cos \theta d\theta$$ Then apply integration by parts to get $$I=-\int_{0}^{\pi/2}\ln\sin\theta d\theta=-\int_{0}^{\pi/2}\ln \cos\theta d\theta=-\frac{1}{2}\int_{0}^{\pi/2}\ln\sin 2\theta d\theta+\frac{1}{2}\int_{0}^{\pi/2}\ln2d\theta$$ The first integral can now be rewritten as below
$$-\frac{1}{4}\int_{0}^{\pi}\ln \sin\theta d\theta=-\frac{1}{4}\int_{0}^{\pi/2}\ln \sin\theta d\theta-\frac{1}{4}\int_{\pi/2}^{\pi}\ln \sin\theta d\theta$$ after a change of variable from $2\theta $ to $\theta$ and then it can be rewritten as $$-\frac{1}{4}\int_{0}^{\pi/2}\ln \sin\theta d\theta-\frac{1}{4}\int_{0}^{\pi/2}\ln \cos\theta d\theta=-\frac{1}{2}\int_{0}^{\pi/2}\ln \sin\theta d\theta=\frac{I}{2}$$ Hence we get $$I=\frac{I}{2}+\frac{\pi}{4}\ln 2\\ \Rightarrow I=\frac{\pi}{2}\ln 2$$
A: Make the substitution $ x \mapsto \sin u $ to arrive at $ \int\limits_0^\frac{\pi}{2} u \cot u \ du $. Proceed with integration by parts to arrive at $ \left[u \ln \sin u \right] _0^{\frac{\pi}{2}}  - \int\limits_0^\frac{\pi}{2} \ln \sin u \ du $. The latter integral is widel known to be $ -\frac{\pi \ln 2}{2} $ and hence the answer is $ \frac{\pi \ln 2}{2} $ as $ \left[u \ln \sin u \right] _0^{\frac{\pi}{2}} = 0 $. 
A: Integrating by parts gives
$$I = \int_0^1 {\frac{{\arcsin x}}{x}dx} = -\int_0^1 {\frac{\ln(x)}{\sqrt{1-x^2}}dx} .$$
Now, consider the integral
$$ F = \int_{0}^{1}\frac{x^{\alpha}}{\sqrt{1-x^2}}dx = \frac{1}{2}\,{\frac {\sqrt {\pi }\,\Gamma  \left( \frac{\alpha}{2}+\frac{1}{2} \right) }{
\Gamma  \left( \frac{\alpha}{2}\,+1 \right) }}, $$
which was evaluated using the beta function. Then our integral $I$ can be evaluate using $F$ as
$$ I = \lim_{\alpha \to 0} \frac{d{F}}{{d\alpha}} =\frac{\ln(2)\pi}{2}.$$
