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Question Number 123488 by peter frank last updated on 25/Nov/20

Answered by MJS_new last updated on 25/Nov/20

x=sech y =(1/(cosh y))=((2e^y )/(e^(2y) +1))  ⇒ y=ln ((1+(√(1−x^2 )))/x) =sech^(−1)  x  ∫sech^(−1)  x dx=∫ln ((1+(√(1−x^2 )))/x) dx=       [by parts]  =xln ((1+(√(1−x^2 )))/x) +∫(dx/( (√(1−x^2 ))))=  =xln ((1+(√(1−x^2 )))/x) +arcsin x +C  [=xsech^(−1)  x +arcsin x +C]

$${x}=\mathrm{sech}\:{y}\:=\frac{\mathrm{1}}{\mathrm{cosh}\:{y}}=\frac{\mathrm{2e}^{{y}} }{\mathrm{e}^{\mathrm{2}{y}} +\mathrm{1}} \\ $$$$\Rightarrow\:{y}=\mathrm{ln}\:\frac{\mathrm{1}+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{{x}}\:=\mathrm{sech}^{−\mathrm{1}} \:{x} \\ $$$$\int\mathrm{sech}^{−\mathrm{1}} \:{x}\:{dx}=\int\mathrm{ln}\:\frac{\mathrm{1}+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{{x}}\:{dx}= \\ $$$$\:\:\:\:\:\left[\mathrm{by}\:\mathrm{parts}\right] \\ $$$$={x}\mathrm{ln}\:\frac{\mathrm{1}+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{{x}}\:+\int\frac{{dx}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}= \\ $$$$={x}\mathrm{ln}\:\frac{\mathrm{1}+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{{x}}\:+\mathrm{arcsin}\:{x}\:+{C} \\ $$$$\left[={x}\mathrm{sech}^{−\mathrm{1}} \:{x}\:+\mathrm{arcsin}\:{x}\:+{C}\right] \\ $$

Answered by Dwaipayan Shikari last updated on 26/Nov/20

sech^(−1) x=t  x=secht  ⇒(2/(e^t +e^(−t) ))⇒e^t +e^(−t) =(2/x)⇒e^(2t) −((2e^t )/x)+1=0  e^t =(1/x)+((√(1−x^2 ))/x) ⇒t=log(((1+(√(1−x^2 )))/x))  ∫log(((1+(√(1−x^2 )))/x))dx=∫log(1+(√(1−x^2 )))−∫logx dx  =xlog(1+(√(1−x^2 )))+∫((x^2 /( (√(1−x^2 ))))/(1+(√(1−x^2 ))))dx−xlogx+x  =xlog(((1+(√(1−x^2 )))/x))+∫(1/( (√(1−x^2 ))))dx   =xlog(((1+(√(1−x^2 )))/x))+sin^(−1) x +C

$${sech}^{−\mathrm{1}} {x}={t} \\ $$$${x}={secht}\:\:\Rightarrow\frac{\mathrm{2}}{{e}^{{t}} +{e}^{−{t}} }\Rightarrow{e}^{{t}} +{e}^{−{t}} =\frac{\mathrm{2}}{{x}}\Rightarrow{e}^{\mathrm{2}{t}} −\frac{\mathrm{2}{e}^{{t}} }{{x}}+\mathrm{1}=\mathrm{0} \\ $$$${e}^{{t}} =\frac{\mathrm{1}}{{x}}+\frac{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{{x}}\:\Rightarrow{t}={log}\left(\frac{\mathrm{1}+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{{x}}\right) \\ $$$$\int{log}\left(\frac{\mathrm{1}+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{{x}}\right){dx}=\int{log}\left(\mathrm{1}+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right)−\int{logx}\:{dx} \\ $$$$={xlog}\left(\mathrm{1}+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right)+\int\frac{\frac{{x}^{\mathrm{2}} }{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}}{\mathrm{1}+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{dx}−{xlogx}+{x} \\ $$$$={xlog}\left(\frac{\mathrm{1}+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{{x}}\right)+\int\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{dx}\: \\ $$$$={xlog}\left(\frac{\mathrm{1}+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{{x}}\right)+{sin}^{−\mathrm{1}} {x}\:+{C} \\ $$

Commented by peter frank last updated on 26/Nov/20

thank you

$$\mathrm{thank}\:\mathrm{you} \\ $$

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