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

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To evaluate the integral (\int_{1}^{2} \frac{1}{\sqrt{x}\sqrt{4-x}} , dx), we can use the substitution method. Let (u = \sqrt{x}). Then, (x = u^2) and (dx = 2u , du). Substituting these into the integral, we get:

[ \int_{1}^{2} \frac{1}{\sqrt{x}\sqrt{4-x}} , dx = \int_{\sqrt{1}}^{\sqrt{2}} \frac{1}{u\sqrt{4-u^2}} \cdot 2u , du ]

[ = 2\int_{1}^{\sqrt{2}} \frac{1}{\sqrt{4-u^2}} , du ]

Now, we can use a trigonometric substitution. Let (u = 2\sin(\theta)), then (du = 2\cos(\theta) , d\theta) and (\sqrt{4-u^2} = 2\cos(\theta)). Substituting these, we get:

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

[ = 2\left[\theta\right]_{\sin^{-1}(1/2)}^{\sin^{-1}(1/\sqrt{2})} = 2\left[\sin^{-1}\left(\frac{1}{\sqrt{2}}\right) - \sin^{-1}\left(\frac{1}{2}\right)\right] ]

[ = 2\left[\frac{\pi}{4} - \frac{\pi}{6}\right] = 2\left(\frac{\pi}{4} - \frac{\pi}{6}\right) = 2\left(\frac{\pi}{12}\right) = \frac{\pi}{6} ]

So, (\int_{1}^{2} \frac{1}{\sqrt{x}\sqrt{4-x}} , dx = \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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