0-1-arcsin-x-x-dx-

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=−∫_0 ^1 ((ln(x))/( (√(1−x^2 )))) x^2 =u=−∫_0 ^1 ((ln(u))/(2(√(1−u)))).(du/(2(√u))) =−(1/4)∫_0 ^1 ln(u)u^(−(1/2)) (1−u)^(−(1/2)) du =−(1/4)∂_a ∫_0 ^1 u^a (1−u)^(−(1/2)) du∣_(=(1/2)) =−(1/4)∂_a β(a,(1/2))∣_(a=(1/2_ )) =−(1/4)(Ψ((1/2))−Ψ(1))β((1/2),(1/2))

$${by}\:{part}=−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left({x}\right)}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }} \\ $$$${x}^{\mathrm{2}} ={u}=−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left({u}\right)}{\mathrm{2}\sqrt{\mathrm{1}−{u}}}.\frac{{du}}{\mathrm{2}\sqrt{{u}}} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left({u}\right){u}^{−\frac{\mathrm{1}}{\mathrm{2}}} \left(\mathrm{1}−{u}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} {du} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{4}}\partial_{{a}} \int_{\mathrm{0}} ^{\mathrm{1}} {u}^{{a}} \left(\mathrm{1}−{u}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} {du}\mid_{=\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{4}}\partial_{{a}} \beta\left({a},\frac{\mathrm{1}}{\mathrm{2}}\right)\mid_{{a}=\frac{\mathrm{1}}{\mathrm{2}_{} }} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{4}}\left(\Psi\left(\frac{\mathrm{1}}{\mathrm{2}}\right)−\Psi\left(\mathrm{1}\right)\right)\beta\left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$

Answered by puissant last updated on 13/Oct/21

Q=∫_0 ^1 ((arcsinx)/x)dx IBP ⇒ Q=−∫_0 ^1 ((lnx)/( (√(1−x^2 ))))dx x=sinu→dx=cosudu ⇒ Q=−∫_0 ^(π/2) ((ln(sinu))/(cosu))cosudu ⇒ Q=−∫_0 ^(π/2) ln(sinu)du ∫_0 ^(π/2) ln(sinx)dx=−(π/2)ln2 ∴∵ Q = ∫_0 ^1 ((arcsinx)/x)dx = (π/2)ln2..

$${Q}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{arcsinx}}{{x}}{dx} \\ $$$$\:{IBP}\:\Rightarrow\:{Q}=−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{lnx}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{dx} \\ $$$${x}={sinu}\rightarrow{dx}={cosudu} \\ $$$$\Rightarrow\:{Q}=−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{ln}\left({sinu}\right)}{{cosu}}{cosudu} \\ $$$$\Rightarrow\:{Q}=−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({sinu}\right){du} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({sinx}\right){dx}=−\frac{\pi}{\mathrm{2}}{ln}\mathrm{2} \\ $$$$ \\ $$$$\:\:\:\:\:\therefore\because\:\:{Q}\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{arcsinx}}{{x}}{dx}\:=\:\frac{\pi}{\mathrm{2}}{ln}\mathrm{2}.. \\ $$

Commented by amin96 last updated on 13/Oct/21

∫_0 ^(π/2) ln(sin(x))dx=^(????) −(π/2)ln2

$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({sin}\left({x}\right)\right){dx}\overset{????} {=}−\frac{\pi}{\mathrm{2}}{ln}\mathrm{2}\:\: \\ $$

Commented by puissant last updated on 13/Oct/21

$${yessssssss}\:{sir}.. \\ $$

Facebook Tweet Pin

Leave a Reply Cancel reply