How do you integrate int sqrt(1-4x-2x^2)∫√1−4x−2x2 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∫(√1−4x−2x2)dx=3√28(2arcsin(√23(1+x))+sin(2arcsin(√23(1+x))))+C

Explanation:

int (sqrt(1-4x-2x^2))dx=sqrt(3)int(sqrt(1-2/3(x+1)^2))dx∫(√1−4x−2x2)dx=√3∫(√1−23(x+1)2)dx

now calling sqrt(2/3)(x+1) = sin y√23(x+1)=siny

sqrt(2/3) dx = cos y dy√23dx=cosydy 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∫(√1−4x−2x2)dx≡√3√32∫cos2ydy=3√2(y2+14sin(2y))+C and after substituting

y = arcsin(sqrt(2/3)(x+1))y=arcsin(√23(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∫(√1−4x−2x2)dx=3√28(2arcsin(√23(1+x))+sin(2arcsin(√23(1+x))))+C

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