How do you integrate #int x^2/sqrt(x^2+1)# by trigonometric substitution?

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Answer 1 · 2018-05-26T15:27:49

#int(x^2*dx)/sqrt(x^2+1)=x/2*sqrt(x^2+1)-1/2sinh^-1x+C#

#int (x^2*dx)/sqrt(x^2+1)#

After using #x=sinhy# and #dx=coshy*dy# transforms, this integral became

#int ((sinhy)^2*coshy*dy)/sqrt((sinhy)^2+1)#

=#int ((sinhy)^2*coshy*dy)/sqrt((coshy)^2)#

=#int ((sinhy)^2*coshy*dy)/coshy#

=#int (sinhy)^2*dy#

=#int (cosh 2y-1)/2*dy#

=#1/4sinh2y-y/2+C#

=#1/4*2sinhy*coshy-y/2+C#

=#1/2sinhy*coshy-y/2+C#

After using #x=sinhy#, #coshy=sqrt(x^2+1)# and #y=sinh^-1x# inverse transforms, I found

#int (x^2*dx)/sqrt(x^2+1)=x/2*sqrt(x^2+1)-1/2sinh^-1x+C#