Question

How do you integrate `int 1/sqrt(-e^(2x)-20e^x-64)dx` using trigonometric substitution?

Answer 1

`- cos^-1((e^x +10)/6) + C`

Explanation:

Complete the square:

`-e^(2x)-20e^x-64 =36 -(e^x+10)^2`

Substitute:

• `e^x +10 = 6 cos u`
• `e^x dx = -(6sin u) du`

Hence:

`int 1/sqrt(-e^(2x)-20e^x-64)dx = int 1/sqrt(36 -(e^x+10)^2)dx`
`= int (-6 sin u)/sqrt(36 -(6cos u)^2) du`
`= int (-6 sin u)/(6sqrt(sin^2 u)) du`
`= int -1du = -u + C = - cos^-1((e^x +10)/6) + C`