Question

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

Answer 1

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

Explanation:

`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`