Question

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

Answer 1

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

Explanation:

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})`