$$ \int\limits_{0}^{1} \frac{\arcsin{x}}{x} dx $$
The answer should be $1.0887...$
I have tried several substituions which didn't work out and also thought of some series expansions
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I think you need to use numerical methods to evaluate this – Vasili Apr 12 '19 at 19:29
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4Change variable to $\theta = \arcsin x$, integrate by part and then apply results here. Your integral evaluates to $\frac{\pi}{2}\log 2 \approx 1.088793045151801$. – achille hui Apr 12 '19 at 19:31
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1@achillehui: Exactly. – Marian G. Apr 12 '19 at 19:34
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Proceed in the following way:
$$I:=\int_{0}^{1}\frac{\arcsin (x)}{x}dx=\int_{0}^{\pi/2}t\cot (t)dt$$
by the substitution $t=\arcsin(x)$. Next, integrating per partes we get
$$I=t\cdot\ln\sin(t)\biggr|_{0}^{\pi/2}-\int_{0}^{\pi/2}\ln (\sin(t))dt.$$
However, the last integral is a well-known one and his value is $-\frac{\pi}{2}\cdot\ln (2)$. Therefore $$I=\frac{\pi}{2}\cdot\ln (2).$$
Marian G.
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1I have reached that integral after substition but I was dumb enough to not integrate by parts :s Thank you! – SADBOYS Apr 12 '19 at 19:37
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