Question Number 156914 by cortano last updated on 17/Oct/21 | ||
Answered by puissant last updated on 17/Oct/21 | ||
$${D}=\int\frac{{arcsin}\left(\frac{\mathrm{1}}{{x}}\right)}{{x}^{\mathrm{5}} }{dx}\:;\:{u}=\frac{\mathrm{1}}{{x}}\:\rightarrow\:{du}=−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }{dx} \\ $$$$\Rightarrow\:{D}=−\int\:\frac{{u}^{\mathrm{5}} {arcsin}\left({u}\right)}{{u}^{\mathrm{2}} }{du}=−\int{u}^{\mathrm{3}} {arcsin}\left({u}\right){du} \\ $$$${IBP}\Rightarrow\:{D}=−\mathrm{3}\left[{u}^{\mathrm{2}} {arcsin}\left({u}\right)\right]+\mathrm{3}\int\frac{{u}^{\mathrm{2}} }{\:\sqrt{\mathrm{1}−{u}^{\mathrm{2}} }}{du} \\ $$$$\Rightarrow\:{D}=−\mathrm{3}{u}^{\mathrm{2}} {arcsin}\left({u}\right)−\mathrm{3}\int\frac{\mathrm{1}−{u}^{\mathrm{2}} −\mathrm{1}}{\:\sqrt{\mathrm{1}−{u}^{\mathrm{2}} }}{du} \\ $$$$\Rightarrow\:{D}=−\mathrm{3}{u}^{\mathrm{2}} {arcsin}\left({u}\right)−\mathrm{3}\int\sqrt{\mathrm{1}−{u}^{\mathrm{2}} }+\mathrm{3}\int\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{u}^{\mathrm{2}} }}{du} \\ $$$$\Rightarrow\:{D}=−\mathrm{3}{u}^{\mathrm{2}} {arcsin}\left({u}\right)−\frac{\mathrm{3}}{\mathrm{2}}\left[{t}+\frac{\mathrm{1}}{\mathrm{2}}{sin}\mathrm{2}{t}\right]+\mathrm{3}{arcsin}\left({u}\right)+{C} \\ $$$$\Rightarrow\:{D}=\frac{−\mathrm{3}}{{x}^{\mathrm{2}} }{arcsin}\left(\frac{\mathrm{1}}{{x}}\right)−\frac{\mathrm{3}}{\mathrm{2}}{arcsin}\left(\frac{\mathrm{1}}{{x}}\right)−\frac{\mathrm{3}}{\mathrm{2}}{sin}\left(\mathrm{2}{arcsin}\left(\frac{\mathrm{1}}{{x}}\right)\right)+\mathrm{3}{arcsin}\left(\frac{\mathrm{1}}{{x}}\right)+{C} \\ $$$$\therefore\because\:{D}=\mathrm{3}\left\{\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right){arcsin}\left(\frac{\mathrm{1}}{{x}}\right)−\frac{\mathrm{1}}{\mathrm{2}}{sin}\left(\mathrm{2}{arcsin}\left(\frac{\mathrm{1}}{{x}}\right)\right)\right\}+{C}.. \\ $$ | ||
Commented by cortano last updated on 17/Oct/21 | ||
$${yes}\:. \\ $$ | ||
Answered by cortano last updated on 17/Oct/21 | ||