Question

How do you integrate `int sqrt(1-4x-2x^2)` using trig substitutions?

Answer 1

`int (sqrt(1-4x-2x^2))dx = 3 sqrt(2)/8(2 arcsin(sqrt(2/3) (1 + x)) + sin(2 arcsin(sqrt(2/3) (1 + x))))+C`

Explanation:

`int (sqrt(1-4x-2x^2))dx=sqrt(3)int(sqrt(1-2/3(x+1)^2))dx`

now calling `sqrt(2/3)(x+1) = sin y`

`sqrt(2/3) dx = cos y dy` and

`int (sqrt(1-4x-2x^2))dx equiv sqrt(3) sqrt(3/2) int cos^2y dy = 3/sqrt(2)(y/2+1/4 sin(2y)) + C` and after substituting

`y = arcsin(sqrt(2/3)(x+1))`

`int (sqrt(1-4x-2x^2))dx = 3 sqrt(2)/8(2 arcsin(sqrt(2/3) (1 + x)) + sin(2 arcsin(sqrt(2/3) (1 + x))))+C`