`I= int (dx)/( sqrt( 4-9 x^(2) ) )`.

Answer

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

### Step 1: Identify the form of the integral
We recognize that the integral resembles the standard form \( \int \frac{dx}{\sqrt{a^2 - x^2}} \), where \( a^2 = 4 \) and \( x^2 = 9x^2 \).

### Step 2: Rewrite the integral
We can factor out constants from the square root:
\[ I = \int \frac{dx}{\sqrt{4 - 9x^2}} = \int \frac{dx}{\sqrt{4(1 - \frac{9}{4}x^2}}} = \int \frac{dx}{2\sqrt{1 - \left(\frac{3}{2}x\right)^2}} \]
This simplifies to:
\[ I = \frac{1}{2} \int \frac{dx}{\sqrt{1 - \left(\frac{3}{2}x\right)^2}} \]

### Step 3: Use substitution
Let \( u = \frac{3}{2}x \). Then, \( du = \frac{3}{2}dx \) or \( dx = \frac{2}{3}du \). Substituting this into the integral gives:
\[ I = \frac{1}{2} \int \frac{\frac{2}{3}du}{\sqrt{1 - u^2}} = \frac{1}{3} \int \frac{du}{\sqrt{1 - u^2}} \]

### Step 4: Integrate using the standard result
The integral \( \int \frac{du}{\sqrt{1 - u^2}} \) is known to be \( \sin^{-1}(u) + C \). Thus,
\[ I = \frac{1}{3} \sin^{-1}(u) + C \]

### Step 5: Substitute back for \( u \)
Now, substituting back \( u = \frac{3}{2}x \):
\[ I = \frac{1}{3} \sin^{-1}\left(\frac{3}{2}x\right) + C \]

### Final Answer
Thus, the final result for the integral is:
\[ I = \frac{1}{3} \sin^{-1}\left(\frac{3}{2}x\right) + C \]