Question Number 228464 by fantastic2 last updated on 17/Apr/26
$$\int{x}\mathrm{sin}^{−\mathrm{1}} {xdx} \\ $$
Commented by fantastic2 last updated on 18/Apr/26
![∫udv=uv−∫vdu here u=sin^(−1) x⇒du=(dx/( (√(1−x^2 )))) and dv=xdx⇒v=(x^2 /2) ∴∫sin^(−1) x×xdx=((x^2 sin^(−1) x)/2)−(1/2)∫(x^2 /( (√(1−x^2 ))))dx −(1/2)∫(x^2 /( (√(1−x^2 ))))dx [x=sin θ⇒dx=cos θdθ] =−(1/2)∫((sin^2 θ)/(cos θ))cos θdθ=−(1/2)∫sin^2 θdθ =−(θ/4)+((sin θcos θ)/4)=−((sin^(−1) x)/4)+((x(√(1−x^2 )))/4) =((x^2 sin^(−1) x)/2)−((sin^(−1) x)/4)+((x(√(1−x^2 )))/4)](https://www.tinkutara.com/question/Q228467.png)
$$\int{udv}={uv}−\int{vdu} \\ $$$${here}\:{u}=\mathrm{sin}^{−\mathrm{1}} {x}\Rightarrow{du}=\frac{{dx}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }} \\ $$$${and}\:{dv}={xdx}\Rightarrow{v}=\frac{{x}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\therefore\int\mathrm{sin}^{−\mathrm{1}} {x}×{xdx}=\frac{{x}^{\mathrm{2}} \mathrm{sin}^{−\mathrm{1}} {x}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{x}^{\mathrm{2}} }{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{dx} \\ $$$$−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{x}^{\mathrm{2}} }{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{dx} \\ $$$$\left[{x}=\mathrm{sin}\:\theta\Rightarrow{dx}=\mathrm{cos}\:\theta{d}\theta\right] \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{sin}\:^{\mathrm{2}} \theta}{\mathrm{cos}\:\theta}\mathrm{cos}\:\theta{d}\theta=−\frac{\mathrm{1}}{\mathrm{2}}\int\mathrm{sin}\:^{\mathrm{2}} \theta{d}\theta \\ $$$$=−\frac{\theta}{\mathrm{4}}+\frac{\mathrm{sin}\:\theta\mathrm{cos}\:\theta}{\mathrm{4}}=−\frac{\mathrm{sin}^{−\mathrm{1}} {x}}{\mathrm{4}}+\frac{{x}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{\mathrm{4}} \\ $$$$=\frac{{x}^{\mathrm{2}} \mathrm{sin}^{−\mathrm{1}} {x}}{\mathrm{2}}−\frac{\mathrm{sin}^{−\mathrm{1}} {x}}{\mathrm{4}}+\frac{{x}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{\mathrm{4}} \\ $$
Answered by Ghisom_ last updated on 17/Apr/26
![∫xarcsin x dx= [by parts] =((x^2 arcsin x)/2)−(1/2)∫(x^2 /( (√(1−x^2 ))))dx= −(1/2)∫(x^2 /( (√(1−x^2 ))))dx= [t=2arcsin x → dx=((√(1−x^2 ))/2)dt] =−(1/8)∫1−cos t dt=−((t−sin t)/8)= =((x(√(1−x^2 )))/4)−((arcsin x)/4) =((2x^2 −1)/4)arcsin x +((x(√(1−x^2 )))/4)+C](https://www.tinkutara.com/question/Q228465.png)
$$\int{x}\mathrm{arcsin}\:{x}\:{dx}= \\ $$$$\:\:\:\:\:\left[\mathrm{by}\:\mathrm{parts}\right] \\ $$$$=\frac{{x}^{\mathrm{2}} \mathrm{arcsin}\:{x}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{x}^{\mathrm{2}} }{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{dx}= \\ $$$$\:\:\:\:\:−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{x}^{\mathrm{2}} }{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{dx}= \\ $$$$\:\:\:\:\:\:\:\:\:\:\left[{t}=\mathrm{2arcsin}\:{x}\:\rightarrow\:{dx}=\frac{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{\mathrm{2}}{dt}\right] \\ $$$$\:\:\:\:\:=−\frac{\mathrm{1}}{\mathrm{8}}\int\mathrm{1}−\mathrm{cos}\:{t}\:{dt}=−\frac{{t}−\mathrm{sin}\:{t}}{\mathrm{8}}= \\ $$$$\:\:\:\:\:=\frac{{x}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{\mathrm{4}}−\frac{\mathrm{arcsin}\:{x}}{\mathrm{4}} \\ $$$$=\frac{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{1}}{\mathrm{4}}\mathrm{arcsin}\:{x}\:+\frac{{x}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{\mathrm{4}}+{C} \\ $$
Commented by fantastic2 last updated on 18/Apr/26
$${sorry}\:{but}\:{i}\:{couldnt}\:{understand}\:{the} \\ $$$${substitution} \\ $$$$\:\:\:\:\:−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{x}^{\mathrm{2}} }{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{dx}=−\frac{\mathrm{1}}{\mathrm{8}}\int\mathrm{1}−\mathrm{cos}\:{t}\:{dt} \\ $$$$\:\:\:\:\: \\ $$$${thank}\:{you} \\ $$
Commented by Ghisom_ last updated on 18/Apr/26
$${I}=−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{x}^{\mathrm{2}} }{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{dx} \\ $$$${t}=\mathrm{2arcsin}\:{x}\:\Leftrightarrow\:{x}=\mathrm{sin}\:\frac{{t}}{\mathrm{2}}\:\rightarrow\:{dx}=\frac{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{\mathrm{2}}{dt} \\ $$$${I}=−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{x}^{\mathrm{2}} }{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}×\frac{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{\mathrm{2}}{dt}= \\ $$$$=−\frac{\mathrm{1}}{\mathrm{4}}\int{x}^{\mathrm{2}} {dt}=−\frac{\mathrm{1}}{\mathrm{4}}\int\mathrm{sin}^{\mathrm{2}} \:\frac{{t}}{\mathrm{2}}\:{dt}= \\ $$$$=−\frac{\mathrm{1}}{\mathrm{4}}\int\frac{\mathrm{1}−\mathrm{cos}\:{t}}{\mathrm{2}}{dt}=−\frac{\mathrm{1}}{\mathrm{8}}\int\mathrm{1}−\mathrm{cos}\:{t}\:{dt}= \\ $$$$=−\frac{{x}}{\mathrm{8}}+\frac{\mathrm{sin}\:{t}}{\mathrm{8}} \\ $$$${t}=\mathrm{2arcsin}\:{x} \\ $$$$\mathrm{sin}\:{t}\:=\mathrm{sin}\:\left(\mathrm{2arcsin}\:{x}\right)\:= \\ $$$$=\mathrm{2cos}\:\left(\mathrm{arcsin}\:{x}\right)\:\mathrm{sin}\:\left(\mathrm{arcsin}\:{x}\right)\:= \\ $$$$=\mathrm{2}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\:{x} \\ $$
Commented by fantastic2 last updated on 18/Apr/26
$${thanks} \\ $$