Question-84496

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 =∫ ((arcsinx)/x^2 )dx by parts u^′ =(1/x^2 ) and v=arcsinx ⇒ I =−((arsinx)/x) −∫−(1/x) ×(1/( (√(1−x^2 ))))dx =−((arcsinx)/x) +∫ (dx/(x(√(1−x^2 )))) cha7gement x=sinθ give ∫ (dx/(x(√(1−x^2 )))) =∫ ((cosθ dθ)/(sinθ.cosθ)) =∫ (dθ/(sinθ)) =_(tan((θ/2))=u) ∫ (1/((2u)/(1+u^2 )))((2du)/(1+u^2 )) =∫ (du/u) =ln∣u∣ +C =ln∣tan((θ/2))∣ +C =ln∣tan(((arcsinx)/2))∣ +C ⇒ I =−((arcsinx)/x) +ln∣tan(((arcsinx)/2))∣ +C

$${I}\:=\int\:\:\frac{{arcsinx}}{{x}^{\mathrm{2}} }{dx}\:\:{by}\:{parts}\:\:{u}^{'} =\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:{and}\:{v}={arcsinx}\:\Rightarrow \\ $$$${I}\:=−\frac{{arsinx}}{{x}}\:−\int−\frac{\mathrm{1}}{{x}}\:×\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{dx} \\ $$$$=−\frac{{arcsinx}}{{x}}\:+\int\:\:\frac{{dx}}{{x}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:\:{cha}\mathrm{7}{gement}\:\:{x}={sin}\theta\:{give} \\ $$$$\int\:\:\frac{{dx}}{{x}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:=\int\:\frac{{cos}\theta\:{d}\theta}{{sin}\theta.{cos}\theta}\:=\int\:\frac{{d}\theta}{{sin}\theta}\:=_{{tan}\left(\frac{\theta}{\mathrm{2}}\right)={u}} \:\:\:\int\:\:\frac{\mathrm{1}}{\frac{\mathrm{2}{u}}{\mathrm{1}+{u}^{\mathrm{2}} }}\frac{\mathrm{2}{du}}{\mathrm{1}+{u}^{\mathrm{2}} } \\ $$$$=\int\:\frac{{du}}{{u}}\:={ln}\mid{u}\mid\:+{C}\:={ln}\mid{tan}\left(\frac{\theta}{\mathrm{2}}\right)\mid\:+{C} \\ $$$$={ln}\mid{tan}\left(\frac{{arcsinx}}{\mathrm{2}}\right)\mid\:+{C}\:\Rightarrow \\ $$$${I}\:=−\frac{{arcsinx}}{{x}}\:+{ln}\mid{tan}\left(\frac{{arcsinx}}{\mathrm{2}}\right)\mid\:+{C} \\ $$$$ \\ $$

Answered by mind is power last updated on 13/Mar/20

u=sin^(−1) (x)⇒ =∫((ucos(u)du)/(sin^2 (u))) by part −(u/(sin(u)))+∫(du/(sin(u)))=−(u/(sin(u)))+∫((sin(u)du)/(1−cos^2 (u))) =−(u/(sin(u)))+∫((sin(u)du)/(2(1−cos(u))))+∫((sin(u)du)/(2(1+cos(u)))) =−(u/(sin(u)))+(1/2)ln(((1−cos(u))/(1+cos(u))))+c =−((sin^− (x))/x)+(1/2)ln(((1−(√(1−x^2 )))/(1+(√(1−x^2 )))))+c

$${u}={sin}^{−\mathrm{1}} \left({x}\right)\Rightarrow \\ $$$$=\int\frac{{ucos}\left({u}\right){du}}{{sin}^{\mathrm{2}} \left({u}\right)} \\ $$$${by}\:{part}\: \\ $$$$−\frac{{u}}{{sin}\left({u}\right)}+\int\frac{{du}}{{sin}\left({u}\right)}=−\frac{{u}}{{sin}\left({u}\right)}+\int\frac{{sin}\left({u}\right){du}}{\mathrm{1}−{cos}^{\mathrm{2}} \left({u}\right)} \\ $$$$=−\frac{{u}}{{sin}\left({u}\right)}+\int\frac{{sin}\left({u}\right){du}}{\mathrm{2}\left(\mathrm{1}−{cos}\left({u}\right)\right)}+\int\frac{{sin}\left({u}\right){du}}{\mathrm{2}\left(\mathrm{1}+{cos}\left({u}\right)\right)} \\ $$$$=−\frac{{u}}{{sin}\left({u}\right)}+\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\frac{\mathrm{1}−{cos}\left({u}\right)}{\mathrm{1}+{cos}\left({u}\right)}\right)+{c} \\ $$$$=−\frac{{sin}^{−} \left({x}\right)}{{x}}+\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\frac{\mathrm{1}−\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{\mathrm{1}+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\right)+{c} \\ $$

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