# How do you evaluate #int dx/sqrt(4-x^2)# from [0,1]?

Evaluate using the second fundamental theorem of calculus.

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To evaluate ( \int_{0}^{1} \frac{dx}{\sqrt{4 - x^2}} ), you can use a trigonometric substitution. Let ( x = 2 \sin(\theta) ), then ( dx = 2 \cos(\theta) , d\theta ), and the limits of integration transform accordingly.

The integral becomes ( \int_{0}^{\frac{\pi}{6}} \frac{2 \cos(\theta) , d\theta}{\sqrt{4 - 4\sin^2(\theta)}} ).

Simplify the integrand using ( \sin^2(\theta) + \cos^2(\theta) = 1 ).

[ \int_{0}^{\frac{\pi}{6}} \frac{2 \cos(\theta) , d\theta}{\sqrt{4 - 4\sin^2(\theta)}} = \int_{0}^{\frac{\pi}{6}} \frac{2 \cos(\theta) , d\theta}{\sqrt{4\cos^2(\theta)}} ]

[ = \int_{0}^{\frac{\pi}{6}} \frac{2 \cos(\theta) , d\theta}{2|\cos(\theta)|} ]

[ = \int_{0}^{\frac{\pi}{6}} d\theta ]

Now, integrate ( d\theta ) with respect to ( \theta ) from ( 0 ) to ( \frac{\pi}{6} ).

[ = \left[\theta\right]_{0}^{\frac{\pi}{6}} ]

[ = \frac{\pi}{6} - 0 ]

[ = \frac{\pi}{6} ]

So, ( \int_{0}^{1} \frac{dx}{\sqrt{4 - x^2}} = \frac{\pi}{6} ).

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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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