# Integrate (1)/(sqrt(1-x^2)) from -1 to 1?

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To integrate ( \frac{1}{\sqrt{1-x^2}} ) from -1 to 1, you can use the definite integral formula for this function, which involves trigonometric substitution.

First, let's express ( \frac{1}{\sqrt{1-x^2}} ) in terms of trigonometric functions by letting ( x = \sin(\theta) ). Then ( dx = \cos(\theta) d\theta ).

[ \int \frac{1}{\sqrt{1-x^2}} , dx = \int \frac{1}{\sqrt{1-\sin^2(\theta)}} , \cos(\theta) , d\theta ] [ = \int \frac{1}{\sqrt{\cos^2(\theta)}} , \cos(\theta) , d\theta ] [ = \int \frac{1}{|\cos(\theta)|} , \cos(\theta) , d\theta ]

Since ( \frac{1}{|\cos(\theta)|} ) is positive on the interval ((-1, 1)), we can drop the absolute value and integrate ( \cos(\theta) ) directly.

[ = \int \cos(\theta) , d\theta ]

The antiderivative of ( \cos(\theta) ) is ( \sin(\theta) ), so:

[ = \sin(\theta) + C ]

Now, we need to evaluate this antiderivative from the lower limit (( -1 )) to the upper limit (( 1 )).

[ \left[ \sin(\theta) \right]_{-1}^{1} ]

Now, find ( \sin(\theta) ) at ( \theta = 1 ) and ( \theta = -1 ):

[ \sin(1) - \sin(-1) ]

[ \sin(1) - (-\sin(1)) ]

[ 2\sin(1) ]

So, the integral of ( \frac{1}{\sqrt{1-x^2}} ) from -1 to 1 is ( 2\sin(1) ).

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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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