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

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Answer 1 · 2016-08-20T16:07:21

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

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

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

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

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

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