Question Number 84496 by bshahid010@gmail.com last updated on 13/Mar/20
Commented by mathmax by abdo last updated on 13/Mar/20
$${I}\:=\int\:\:\frac{{arcsinx}}{{x}^{\mathrm{2}} }{dx}\:\:{by}\:{parts}\:\:{u}^{'} =\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:{and}\:{v}={arcsinx}\:\Rightarrow \\ $$$${I}\:=−\frac{{arsinx}}{{x}}\:−\int−\frac{\mathrm{1}}{{x}}\:×\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{dx} \\ $$$$=−\frac{{arcsinx}}{{x}}\:+\int\:\:\frac{{dx}}{{x}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:\:{cha}\mathrm{7}{gement}\:\:{x}={sin}\theta\:{give} \\ $$$$\int\:\:\frac{{dx}}{{x}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:=\int\:\frac{{cos}\theta\:{d}\theta}{{sin}\theta.{cos}\theta}\:=\int\:\frac{{d}\theta}{{sin}\theta}\:=_{{tan}\left(\frac{\theta}{\mathrm{2}}\right)={u}} \:\:\:\int\:\:\frac{\mathrm{1}}{\frac{\mathrm{2}{u}}{\mathrm{1}+{u}^{\mathrm{2}} }}\frac{\mathrm{2}{du}}{\mathrm{1}+{u}^{\mathrm{2}} } \\ $$$$=\int\:\frac{{du}}{{u}}\:={ln}\mid{u}\mid\:+{C}\:={ln}\mid{tan}\left(\frac{\theta}{\mathrm{2}}\right)\mid\:+{C} \\ $$$$={ln}\mid{tan}\left(\frac{{arcsinx}}{\mathrm{2}}\right)\mid\:+{C}\:\Rightarrow \\ $$$${I}\:=−\frac{{arcsinx}}{{x}}\:+{ln}\mid{tan}\left(\frac{{arcsinx}}{\mathrm{2}}\right)\mid\:+{C} \\ $$$$ \\ $$
Answered by mind is power last updated on 13/Mar/20
$${u}={sin}^{−\mathrm{1}} \left({x}\right)\Rightarrow \\ $$$$=\int\frac{{ucos}\left({u}\right){du}}{{sin}^{\mathrm{2}} \left({u}\right)} \\ $$$${by}\:{part}\: \\ $$$$−\frac{{u}}{{sin}\left({u}\right)}+\int\frac{{du}}{{sin}\left({u}\right)}=−\frac{{u}}{{sin}\left({u}\right)}+\int\frac{{sin}\left({u}\right){du}}{\mathrm{1}−{cos}^{\mathrm{2}} \left({u}\right)} \\ $$$$=−\frac{{u}}{{sin}\left({u}\right)}+\int\frac{{sin}\left({u}\right){du}}{\mathrm{2}\left(\mathrm{1}−{cos}\left({u}\right)\right)}+\int\frac{{sin}\left({u}\right){du}}{\mathrm{2}\left(\mathrm{1}+{cos}\left({u}\right)\right)} \\ $$$$=−\frac{{u}}{{sin}\left({u}\right)}+\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\frac{\mathrm{1}−{cos}\left({u}\right)}{\mathrm{1}+{cos}\left({u}\right)}\right)+{c} \\ $$$$=−\frac{{sin}^{−} \left({x}\right)}{{x}}+\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\frac{\mathrm{1}−\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{\mathrm{1}+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\right)+{c} \\ $$