Question

How do you integrate `int e^(2x)/sqrt(e^(2x) -36)dx` using trigonometric substitution?

Answer 1

`sqrt(e^(2x)-36) + C`

Explanation:

We don't really need substitution since we have a perfect form
`int (f'(x))/(2sqrt(f(x))) ->sqrt(f(x))`
but here you go:

• `e^x = 6cosh u`
• `e^x dx = (6 sinh u) du`

Hence:

`e^(2x) - 36 = (6cosh u)^2 - 36 = 36(cosh^2 u - 1)`
`36sinh^2 u= (6sinh u)^2`

So we have:

`int e^(2x)/sqrt(e^(2x) -36)dx = int e^x/sqrt(e^(2x) -36) (e^xdx) = int (6coshu)/(6 sinh u) (6sinhu du) = int 6coshu du = 6sinh u + C`

To substitute back we use:
`sinhu = sqrt(cosh^2u -1 ) = sqrt(e^(2x)/36-1) = sqrt(e^(2x)-36)/6`

Finally:

`int e^(2x)/sqrt(e^(2x) -36)dx=sqrt(e^(2x)-36) + C`