Question Number 84666 by jagoll last updated on 14/Mar/20
$$\int\:{x}\:\mathrm{sin}^{−\mathrm{1}} \left({x}\right)\:{dx}\: \\ $$
Commented by mathmax by abdo last updated on 15/Mar/20
$${A}\:=\int\:{xarcsinx}\:{dx}\:\:{by}\:{parts}\:\:{u}^{'} ={x}\:{and}\:{v}={arcsinx}\:\Rightarrow \\ $$$${A}\:=\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:{arcsinx}\:−\int\:\:\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\frac{{dx}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:\:{we}\:{have} \\ $$$$\int\:\:\frac{{x}^{\mathrm{2}} {dx}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:=\int\:\frac{{x}^{\mathrm{2}} −\mathrm{1}\:+\mathrm{1}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{dx}\:=\int\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\:+\int\:\frac{{dx}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }} \\ $$$$=−\int\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }{dx}\:+{arcsinx}\:+{C}\:\:{but} \\ $$$$\int\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }{dx}\:=_{{x}={sint}} \:\:\:\int{cost}\:{cost}\:{dt}\:=\int\:\frac{\mathrm{1}+{cos}\left(\mathrm{2}{t}\right)}{\mathrm{2}}{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{t}\:+\frac{\mathrm{1}}{\mathrm{4}}{sin}\left(\mathrm{2}{t}\right)\:=\frac{\mathrm{1}}{\mathrm{2}}{t}\:+\frac{\mathrm{1}}{\mathrm{2}}{sint}\:{cost}\:=\frac{\mathrm{1}}{\mathrm{2}}{arcsinx}\:+\frac{\mathrm{1}}{\mathrm{2}}{x}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} } \\ $$$$\Rightarrow{A}\:=\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}} \:{arcsinx}\:−\frac{\mathrm{1}}{\mathrm{2}}\left(\:{arcsinx}\:\:−\frac{\mathrm{1}}{\mathrm{2}}{arcsinx}\:−\frac{\mathrm{1}}{\mathrm{2}}{x}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right)\:+{C} \\ $$$$=\frac{{x}^{\mathrm{2}} }{\mathrm{2}}{arcsinx}\:−\frac{\mathrm{1}}{\mathrm{4}}\:{arcsinx}\:+\frac{\mathrm{1}}{\mathrm{4}}{x}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\:+{C} \\ $$$$=\left(\frac{{x}^{\mathrm{2}} }{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{4}}\right){arcsinx}\:−\frac{{x}}{\mathrm{4}}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\:+{C} \\ $$$$ \\ $$
Commented by mathmax by abdo last updated on 15/Mar/20
$${A}\:=\left(\frac{{x}^{\mathrm{2}} }{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{4}}\right){arcsinx}\:+\frac{{x}}{\mathrm{4}}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\:+{C} \\ $$
Answered by MJS last updated on 15/Mar/20
![∫xarcsin x dx= by parts u=arcsin x → u′=(1/( (√(1−x^2 )))) v′=x → v=(1/2)x^2 =(1/2)x^2 arcsin x −(1/2)∫(x^2 /( (√(1−x^2 ))))dx= ∫(x^2 /( (√(1−x^2 ))))dx= [t=arcsin x → dx=cos t dt] =∫sin^2 t dt=(1/4)(2t−sin 2t)= =(1/2)(arcsin x −x(√(1−x^2 ))) =(1/4)((2x^2 −1)arcsin x +x(√(1−x^2 ))) +C](https://www.tinkutara.com/question/Q84671.png)
$$\int{x}\mathrm{arcsin}\:{x}\:{dx}= \\ $$$$\:\:\:\:\:\mathrm{by}\:\mathrm{parts} \\ $$$$\:\:\:\:\:{u}=\mathrm{arcsin}\:{x}\:\rightarrow\:{u}'=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }} \\ $$$$\:\:\:\:\:{v}'={x}\:\rightarrow\:{v}=\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}} \mathrm{arcsin}\:{x}\:−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{x}^{\mathrm{2}} }{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{dx}= \\ $$$$ \\ $$$$\:\:\:\:\:\int\frac{{x}^{\mathrm{2}} }{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{dx}= \\ $$$$\:\:\:\:\:\:\:\:\:\:\left[{t}=\mathrm{arcsin}\:{x}\:\rightarrow\:{dx}=\mathrm{cos}\:{t}\:{dt}\right] \\ $$$$\:\:\:\:\:=\int\mathrm{sin}^{\mathrm{2}} \:{t}\:{dt}=\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{2}{t}−\mathrm{sin}\:\mathrm{2}{t}\right)= \\ $$$$\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{arcsin}\:{x}\:−{x}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right) \\ $$$$ \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left(\left(\mathrm{2}{x}^{\mathrm{2}} −\mathrm{1}\right)\mathrm{arcsin}\:{x}\:+{x}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right)\:+{C} \\ $$
Commented by jagoll last updated on 15/Mar/20
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$