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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

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

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

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

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

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

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