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Question Number 156616 by amin96 last updated on 13/Oct/21

$${\int }_0^1\frac{arcsin\left(x\right)}{x}dx=?$$

Answered by mindispower last updated on 13/Oct/21

$$bypart=-{\int }_0^1\frac{ln\left(x\right)}{\sqrt{1-x^2}}$$$$x^2=u=-{\int }_0^1\frac{ln\left(u\right)}{2\sqrt{1-u}}.\frac{du}{2\sqrt{u}}$$$$=-\frac{1}{4}{\int }_0^{1}ln\left(u\right)u^{-\frac{1}{2}}{(1-u)}^{-\frac{1}{2}}du$$$$=-\frac{1}{4}{\partial }_a{\int }_0^1{u}^a{(1-u)}^{-\frac{1}{2}}du{\mid }_{=\frac{1}{2}}$$$$=-\frac{1}{4}{\partial }_a\beta (a,\frac{1}{2}){\mid }_{a=\frac{1}{2_{}}}$$$$=-\frac{1}{4}(\mathrm{\Psi }\left(\frac{1}{2}\right)-\mathrm{\Psi }\left(1\right))\beta (\frac{1}{2},\frac{1}{2})$$

Answered by puissant last updated on 13/Oct/21

$$Q={\int }_0^1\frac{arcsinx}{x}dx$$$$IBP\Rightarrow Q=-{\int }_0^1\frac{lnx}{\sqrt{1-x^2}}dx$$$$x=sinu\to dx=cosudu$$$$\Rightarrow Q=-{\int }_0^{\frac{\pi }{2}}\frac{ln\left(sinu\right)}{cosu}cosudu$$$$\Rightarrow Q=-{\int }_0^{\frac{\pi }{2}}ln\left(sinu\right)du$$$${\int }_0^{\frac{\pi }{2}}ln\left(sinx\right)dx=-\frac{\pi }{2}ln2$$$$$$$$\therefore \because Q={\int }_0^1\frac{arcsinx}{x}dx=\frac{\pi }{2}ln2..$$

Commented by amin96 last updated on 13/Oct/21

$${\int }_0^{\frac{\pi }{2}}ln\left(sin\left(x\right)\right)dx\stackrel{????}{=}-\frac{\pi }{2}ln2$$

Commented by puissant last updated on 13/Oct/21

$$yesssssssssir..$$

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