Question Number 156616 by amin96 last updated on 13/Oct/21 | ||
$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{arcsin}\left({x}\right)}{{x}}{dx}=? \\ $$ | ||
Answered by mindispower last updated on 13/Oct/21 | ||
$${by}\:{part}=−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left({x}\right)}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }} \\ $$$${x}^{\mathrm{2}} ={u}=−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left({u}\right)}{\mathrm{2}\sqrt{\mathrm{1}−{u}}}.\frac{{du}}{\mathrm{2}\sqrt{{u}}} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left({u}\right){u}^{−\frac{\mathrm{1}}{\mathrm{2}}} \left(\mathrm{1}−{u}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} {du} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{4}}\partial_{{a}} \int_{\mathrm{0}} ^{\mathrm{1}} {u}^{{a}} \left(\mathrm{1}−{u}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} {du}\mid_{=\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{4}}\partial_{{a}} \beta\left({a},\frac{\mathrm{1}}{\mathrm{2}}\right)\mid_{{a}=\frac{\mathrm{1}}{\mathrm{2}_{} }} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{4}}\left(\Psi\left(\frac{\mathrm{1}}{\mathrm{2}}\right)−\Psi\left(\mathrm{1}\right)\right)\beta\left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$ | ||
Answered by puissant last updated on 13/Oct/21 | ||
$${Q}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{arcsinx}}{{x}}{dx} \\ $$$$\:{IBP}\:\Rightarrow\:{Q}=−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{lnx}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{dx} \\ $$$${x}={sinu}\rightarrow{dx}={cosudu} \\ $$$$\Rightarrow\:{Q}=−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{ln}\left({sinu}\right)}{{cosu}}{cosudu} \\ $$$$\Rightarrow\:{Q}=−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({sinu}\right){du} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({sinx}\right){dx}=−\frac{\pi}{\mathrm{2}}{ln}\mathrm{2} \\ $$$$ \\ $$$$\:\:\:\:\:\therefore\because\:\:{Q}\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{arcsinx}}{{x}}{dx}\:=\:\frac{\pi}{\mathrm{2}}{ln}\mathrm{2}.. \\ $$ | ||
Commented by amin96 last updated on 13/Oct/21 | ||
$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({sin}\left({x}\right)\right){dx}\overset{????} {=}−\frac{\pi}{\mathrm{2}}{ln}\mathrm{2}\:\: \\ $$ | ||
Commented by puissant last updated on 13/Oct/21 | ||
$${yessssssss}\:{sir}.. \\ $$ | ||