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 these steps:

### Step 1: Rewrite the Integral
We start by rewriting 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 2: Identify the Form
Notice that the integral now resembles the standard form:
\[ \int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\left(\frac{x}{a}\right) + C \]
where \( a = \frac{4}{3} \) (since \( \frac{16}{9} = \left(\frac{4}{3}\right)^2 \)).

### Step 3: Apply the Standard Integral
Now applying the standard integral formula:
\[ I = \frac{1}{3} \left( \sin^{-1}\left(\frac{x}{\frac{4}{3}}\right) + C \right) \]
This simplifies to:
\[ I = \frac{1}{3} \sin^{-1}\left(\frac{3x}{4}\right) + C \]

### Final Answer
Thus, the evaluated integral is:
\[ I = \frac{1}{3} \sin^{-1}\left(\frac{3x}{4}\right) + C \]

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