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So my integral is $$\int_{0}^{1}\frac{\sin^{-1}(x)}{x}$$ To avoid confusion let me re-write the integral as $$\mathcal I = \int_0^1 \frac{\arcsin(x)}{x}$$ I started off with a trig-substitution that is let $x = \sin(t)$ and $t = \arcsin(x)$ which means that $dx = \cos(t) dt$ So our integrand becomes $$\mathcal I = \int_0^{\frac{\pi}{2}} \frac{t}{\sin(t)} \cos(t) dt\tag{Bounds have changed}$$ $$= \int_0^{\frac{\pi}{2}} t\space\cot(t) dt$$ Then using Integration by Parts,$\space$$u = t$ $\implies du = dt$ and $dv = \cot(t)$ $\implies v = \ln(\sin(t))$ So our integrand thus becomes, $= t\space\ln(\sin(t))$ from $0$ to $\frac{\pi}{2}$ $$-\int_0^{\frac{\pi}{2}} \ln(sin(t))dt\tag{t*ln(sin(t)) = 0}$$ From here, I don't know how to proceed further. Any help/hint is appreciated :)

Thanks in advance