Home > Calculus > Integration by Trigonometric Substitution

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

Answer 1

Samantha Adkison

#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#

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Aurora Alvarado

To integrate (\int \sqrt{1-4x-2x^2} , dx) using trigonometric substitution, first complete the square inside the square root. After that, make a trigonometric substitution by letting (x = -\frac{1}{2} + \frac{1}{2}\sin(\theta)) or (x = -\frac{1}{2} - \frac{1}{2}\sin(\theta)) depending on the form of the quadratic expression. Then, express the integral in terms of (\theta), and use trigonometric identities to simplify it. Finally, integrate with respect to (\theta) and substitute back in terms of (x).

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.