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

We can use a trigonometric substitution to evaluate this integral.

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Recalling that:

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To find the integral of ( \frac{1}{\sqrt{1-4x^2}} ) with respect to ( x ), you can use trigonometric substitution. Let ( x = \frac{1}{2}\sin(\theta) ). Then, ( dx = \frac{1}{2}\cos(\theta) d\theta ). Substitute these expressions into the integral and simplify. You should end up with ( \int \frac{1}{\sqrt{1-4x^2}} dx = \frac{1}{2} \int \frac{1}{\cos(\theta)} d\theta ). This simplifies further to ( \frac{1}{2} \int \sec(\theta) d\theta ). The integral of ( \sec(\theta) ) can be found using the natural logarithm function. Therefore, the final result is ( \frac{1}{2} \ln|\sec(\theta) + \tan(\theta)| + C ), where ( C ) is the constant of integration. Finally, substitute back ( \theta ) in terms of ( x ) to get the integral in terms of ( x ).

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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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