How do you integrate #int 1/sqrt(-e^(2x)+12e^x-11)dx# using trigonometric substitution?

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Answer 1 · 2018-03-26T21:11:34

#int (dx)/sqrt(-e^(2x)+12e^x-11)=-sqrt11/11arcsin((11e^(-x)-6)/5)+C#

#int (dx)/sqrt(-e^(2x)+12e^x-11)#

=#int (e^(-x)*dx)/sqrt(-11e^(-2x)+12e^(-x)-1)#

=#sqrt11int (e^(-x)*dx)/sqrt(-121e^(-2x)+132e^(-x)-11)#

=#sqrt11int (e^(-x)*dx)/sqrt(5^2-(11e^(-x)-6)^2)#

=#-sqrt11/11int (-11e^(-x)*dx)/sqrt(5^2-(11e^(-x)-6)^2)#

After using #11e^(-x)-6=5siny# and #-11e^(-x)*dx=5cosy*dy# transforms, this integral became

#-sqrt11/11int (5cosy*dy)/sqrt(5^2-(5siny)^2)#

=#-sqrt11/11int (5cosy*dy)/(5cosy)#

=#-sqrt11/11int dy#

=#-sqrt11/11y+C#

After using #11e^(-x)-6=5siny# and #y=arcsin((11e^(-x)-6)/5)# inverse transforms, I found

#int (dx)/sqrt(-e^(2x)+12e^x-11)=-sqrt11/11arcsin((11e^(-x)-6)/5)+C#