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

Substitute:

Then:

so:

and undoing the substitution:

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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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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.

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.

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.

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