How do you integrate #int e^-x/(9e^(-2x)+1)^(3/2)# by trigonometric substitution?

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Answer 1 · 2016-07-20T11:13:45

#-e^{-x}/sqrt(1+9e^{-2x})#

Making #3 e^{-x} = i cos(y)# then
#-3 e^{-x}dx = -i sin(y)dy#

Substituting in the integral we have

#int (e^-x dx)/(9e^(-2x)+1)^(3/2) equiv 1/3int (i sin(y) dy)/(1-cos^2(y))^{3/2} = i/3int (dy)/sin^2(y) =- i/3cot(y)#

but

#-i/3cot(y)=-i/3cos(y)/sqrt(1-cos^2(y)) = -i/3(-3ie^{-x})/sqrt(1+9e^{-2x})#

and finally

#int (e^-x dx)/(9e^(-2x)+1)^(3/2)=-e^{-x}/sqrt(1+9e^{-2x})#