Evaluate the following integrals:
`int frac{dx}{sqrt(16-9x^2)}`

Answer

To evaluate the integral

\[ I = \int \frac{dx}{\sqrt{16 - 9x^2}}, \]

we will follow a series of steps.

### Step 1: Rewrite the Integral
First, we can factor out the constant from the square root in the denominator. Notice that we can express \(16\) as \(\frac{16}{9} \cdot 9\):

\[ I = \int \frac{dx}{\sqrt{16 - 9x^2}} = \int \frac{dx}{\sqrt{9 \left(\frac{16}{9} - x^2\right)}}. \]

### Step 2: Simplify the Square Root
Now, we can take the square root of \(9\) out of the integral:

\[ I = \int \frac{dx}{3\sqrt{\frac{16}{9} - x^2}} = \frac{1}{3} \int \frac{dx}{\sqrt{\frac{16}{9} - x^2}}. \]

### Step 3: Recognize the Standard Form
Next, we recognize that \(\frac{16}{9}\) can be rewritten as \(\left(\frac{4}{3}\right)^2\):

\[ I = \frac{1}{3} \int \frac{dx}{\sqrt{\left(\frac{4}{3}\right)^2 - x^2}}. \]

This integral is in the standard form of

\[ \int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\left(\frac{x}{a}\right) + C, \]

where \(a = \frac{4}{3}\).

### Step 4: Apply the Standard Integral
Using the standard integral, we have:

\[ I = \frac{1}{3} \cdot \sin^{-1}\left(\frac{x}{\frac{4}{3}}\right) + C. \]

### Step 5: Simplify the Result
Now, we simplify the expression:

\[ I = \frac{1}{3} \sin^{-1}\left(\frac{3x}{4}\right) + C. \]

### Final Answer
Thus, the evaluated integral is:

\[ I = \frac{1}{3} \sin^{-1}\left(\frac{3x}{4}\right) + C. \]

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