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

1 Answer
Cesareo R.
Aug 20, 2016

#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#

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