`int (dx)/(sqrt(16-9x^(2)))`

Explanation

To solve the integral \( \int \frac{dx}{\sqrt{16 - 9x^2}} \), we can follow these steps:

### Step 1: Rewrite the integral
We start with the integral:
\[ I = \int \frac{dx}{\sqrt{16 - 9x^2}} \]

### Step 2: Factor out the coefficient of \(x^2\)
Notice that we can factor out the 9 from the expression under the square root:
\[ I = \int \frac{dx}{\sqrt{16 - 9x^2}} = \int \frac{dx}{\sqrt{9\left(\frac{16}{9} - x^2\right)}} \]
This simplifies to:
\[ I = \int \frac{dx}{3\sqrt{\frac{16}{9} - x^2}} = \frac{1}{3} \int \frac{dx}{\sqrt{\frac{16}{9} - x^2}} \]

### Step 3: Rewrite the expression under the square root
Next, we rewrite \( \frac{16}{9} \) as \( \left(\frac{4}{3}\right)^2 \):
\[ I = \frac{1}{3} \int \frac{dx}{\sqrt{\left(\frac{4}{3}\right)^2 - x^2}} \]

### Step 4: Use the standard integral formula
We can now use the standard integral formula:
\[ \int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\left(\frac{x}{a}\right) + C \]
where \( a = \frac{4}{3} \).

### Step 5: Apply the formula
Applying the formula, we have:
\[ I = \frac{1}{3} \sin^{-1}\left(\frac{x}{\frac{4}{3}}\right) + C \]
This simplifies to:
\[ I = \frac{1}{3} \sin^{-1}\left(\frac{3x}{4}\right) + C \]

### Final Answer
Thus, the final result of the integral is:
\[ \int \frac{dx}{\sqrt{16 - 9x^2}} = \frac{1}{3} \sin^{-1}\left(\frac{3x}{4}\right) + C \]