Evaluate the following integrals:
`int(1)/(sqrt(9-4x^(2)))dx`

Answer

To evaluate the integral \( \int \frac{1}{\sqrt{9 - 4x^2}} \, dx \), we can follow these steps:

### Step 1: Identify the form of the integral
We notice that the integral has the form \( \int \frac{1}{\sqrt{a^2 - u^2}} \, du \), where \( a^2 = 9 \) and \( u^2 = 4x^2 \).

### Step 2: Rewrite the expression
We can rewrite \( 9 - 4x^2 \) as \( 3^2 - (2x)^2 \). This gives us:
\[ \int \frac{1}{\sqrt{3^2 - (2x)^2}} \, dx \]

### Step 3: Use the standard integral formula
We know from integral calculus that:
\[ \int \frac{1}{\sqrt{a^2 - u^2}} \, du = \sin^{-1} \left( \frac{u}{a} \right) + C \]
In our case, \( a = 3 \) and \( u = 2x \).

### Step 4: Adjust for the coefficient of \( x \)
Since we have \( u = 2x \), we need to account for the derivative of \( u \) with respect to \( x \). The derivative \( \frac{du}{dx} = 2 \), so we need to multiply our integral by \( \frac{1}{2} \) to account for this. Thus, we have:
\[ \int \frac{1}{\sqrt{3^2 - (2x)^2}} \, dx = \frac{1}{2} \sin^{-1} \left( \frac{2x}{3} \right) + C \]

### Final Answer
Putting it all together, we get:
\[ \int \frac{1}{\sqrt{9 - 4x^2}} \, dx = \frac{1}{2} \sin^{-1} \left( \frac{2x}{3} \right) + C \]

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