Question Number 163886 by zakirullah last updated on 11/Jan/22
$${Evaluate}\:{the}\:{following}\:{by}\:{using}\: \\ $$$$\:{integration}\:{by}\:{parts}\:{formula}: \\ $$$$\int{xsin}^{−\mathrm{1}} \left({x}\right){dx} \\ $$
Commented by Ar Brandon last updated on 11/Jan/22
$${u}\left({x}\right)=\mathrm{sin}^{−\mathrm{1}} \left({x}\right)\:\:,\:\:\:{v}'\left({x}\right)={x} \\ $$
Commented by zakirullah last updated on 11/Jan/22
$${dear}\:\boldsymbol{{S}}{ir}!\:{need}\:{solution}? \\ $$
Answered by Ar Brandon last updated on 11/Jan/22
$$\int{x}\mathrm{sin}^{−\mathrm{1}} \left({x}\right){dx}=\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}} \mathrm{sin}^{−\mathrm{1}} \left({x}\right)−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{x}^{\mathrm{2}} }{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}} \mathrm{sin}^{−\mathrm{1}} \left({x}\right)−\frac{\mathrm{1}}{\mathrm{2}}\left({x}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }−\int\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }{dx}\right) \\ $$$$\int\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }{dx}=\frac{\mathrm{1}}{\mathrm{2}}\int\left(\mathrm{1}+\mathrm{cos2}\vartheta\right){d}\vartheta\:,\:{x}=\mathrm{sin}\vartheta \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{sin}^{−\mathrm{1}} \left({x}\right)+{x}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right)+{C} \\ $$$$\Rightarrow\int{x}\mathrm{sin}^{−\mathrm{1}} \left({x}\right){dx}=\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}} \mathrm{sin}^{−\mathrm{1}} \left({x}\right)−\frac{\mathrm{1}}{\mathrm{4}}{x}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{4}}\mathrm{sin}^{−\mathrm{1}} \left({x}\right)+{C} \\ $$
Commented by zakirullah last updated on 12/Jan/22
