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How do you find the integral of #int 3/sqrt(1-4x^2)dx#?

Answer 1

Samantha Adkison

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

Substitute:

#x =1/2 sint#

#dx = 1/2costdt#

with #t in (-pi/2,pi/2)# as the integrand is defined only for #abs (2x) < 1#.

Then:

#int 3/sqrt(1-4x^2)dx = 3/2 int (costdt)/sqrt(1-4(1/2sint)^2) #

#int 3/sqrt(1-4x^2)dx = 3/2 int (costdt)/sqrt(1-sin^2t) #

Now, for #t in (-pi/2,pi/2)#:

#sqrt(1-sin^2t) = cost#

so:

#int 3/sqrt(1-4x^2)dx = 3/2 int (costdt)/cost#

#int 3/sqrt(1-4x^2)dx = 3/2 intdt#

#int 3/sqrt(1-4x^2)dx = (3t)/2 + C#

and undoing the substitution:

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

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Answer 2

William Abdalla

To find the integral of ( \int \frac{3}{\sqrt{1-4x^2}} , dx ), use the trigonometric substitution method. Let ( x = \frac{1}{2} \sin(\theta) ), then ( dx = \frac{1}{2} \cos(\theta) , d\theta ). Substitute these into the integral and simplify using trigonometric identities. You will end up with ( \int \frac{3}{\sqrt{1-4x^2}} , dx = \int \frac{3}{\sqrt{1-\sin^2(\theta)}} \cdot \frac{1}{2} \cos(\theta) , d\theta ). Simplify further and integrate to get the final answer.

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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.