# How do you integrate # (1-x^2)^.5#?

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To integrate ( \sqrt{1 - x^2} ), you can use a trigonometric substitution. Let ( x = \sin(\theta) ). Then, ( dx = \cos(\theta) , d\theta ).

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

Using the Pythagorean identity ( \cos^2(\theta) = 1 - \sin^2(\theta) ), we have ( \cos(\theta) = \sqrt{1 - \sin^2(\theta)} ).

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

Now, recall that ( \cos^2(\theta) = \frac{1}{2}(1 + \cos(2\theta)) ).

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

Integrating term by term, you get:

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

Finally, substitute back ( x = \sin(\theta) ) and simplify the expression:

[ \int \sqrt{1 - x^2} , dx = \frac{1}{2} (\arcsin(x) + x\sqrt{1-x^2}) + C ]

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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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