# How do you evaluate the integral #int 1/(sqrt(4-x^2)dx# from 0 to 2?

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To evaluate the integral ( \int_0^2 \frac{1}{\sqrt{4-x^2}} , dx), we can use the substitution method. Let ( x = 2\sin(\theta) ), then ( dx = 2\cos(\theta) , d\theta ) and the limits of integration become ( \theta = \sin^{-1}(0) = 0 ) and ( \theta = \sin^{-1}(1) = \frac{\pi}{2} ).

[ \int_0^2 \frac{1}{\sqrt{4-x^2}} , dx = \int_0^{\frac{\pi}{2}} \frac{1}{\sqrt{4-(2\sin(\theta))^2}} \cdot 2\cos(\theta) , d\theta ]

Simplify the expression under the square root and integrate:

[ = \int_0^{\frac{\pi}{2}} \frac{1}{\sqrt{4-4\sin^2(\theta)}} \cdot 2\cos(\theta) , d\theta ]

[ = \int_0^{\frac{\pi}{2}} \frac{1}{\sqrt{4\cos^2(\theta)}} \cdot 2\cos(\theta) , d\theta ]

[ = \int_0^{\frac{\pi}{2}} \frac{1}{2\cos(\theta)} \cdot 2\cos(\theta) , d\theta ]

[ = \int_0^{\frac{\pi}{2}} d\theta ]

Now, integrating (d\theta) over the interval (0) to (\frac{\pi}{2}) gives:

[ = \left[\theta\right]_0^{\frac{\pi}{2}} ]

[ = \frac{\pi}{2} - 0 ]

[ = \frac{\pi}{2} ]

So, ( \int_0^2 \frac{1}{\sqrt{4-x^2}} , dx = \frac{\pi}{2} ).

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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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