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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

p_square
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    You have been a bit cavalier with the boundary terms, but with limit thrown in you're right. This is a standard definite integral. Try integrating from $0$ to $\pi$ and letting $t=2u$. – Ted Shifrin Jun 28 '21 at 17:10
  • @TedShifrin I found the bounds by substituting value of $x=0, 1$ and got my final bounds as $0$ and $\frac{\pi}{2}$. So can you please clarify how am I being cavalier with the boundary terms. Thanks :) – p_square Jun 28 '21 at 17:16
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    Yes, I'm suggesting that you instead consider the integral from $0$ to $\pi$; that will make the substitution work out cleverly. You cannot plug $t=0$ into the function $t\ln(\sin t)$, right? – Ted Shifrin Jun 28 '21 at 17:22
  • @TedShifrin Oh yeah... Now I understood my mistake. Thank you so much :) – p_square Jun 28 '21 at 17:49
  • @Algebrology see here in this link I have computed the original integral: https://twitter.com/AlbahariRicardo/status/1403091258298572800 and here another method for the second integral: https://mymatexnotes.blogspot.com/2021/06/evaluating-some-log-trig-integrals.html – Ricardo770 Jun 28 '21 at 19:03

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