`int(dx)/(sqrt(16-4x^(2)))=?`

Options

A. `(1)/(2) sin ^(-1)""(x)/(2)+C`
B. `(1)/(4) sin ^(-1)""(x)/(2)+C`
C. `(1)/(2) sin ^(-1)""(x)/(4)+C`
D. None of these

Explanation

To solve the integral \( \int \frac{dx}{\sqrt{16 - 4x^2}} \), we can follow these steps:

### Step 1: Rewrite the Integral
We start with the integral:
\[ I = \int \frac{dx}{\sqrt{16 - 4x^2}} \]
To simplify the expression under the square root, we can factor out the constant from the square root.

### Step 2: Factor Out the Constant
Notice that \( 16 - 4x^2 \) can be rewritten as:
\[ 16 - 4x^2 = 4(4 - x^2) \]
Thus, we can rewrite the integral as:
\[ I = \int \frac{dx}{\sqrt{4(4 - x^2)}} \]
This simplifies to:
\[ I = \int \frac{dx}{2\sqrt{4 - x^2}} = \frac{1}{2} \int \frac{dx}{\sqrt{4 - x^2}} \]

### Step 3: Use the Standard Integral Formula
The integral \( \int \frac{dx}{\sqrt{a^2 - x^2}} \) is a standard integral that evaluates to \( \sin^{-1} \left( \frac{x}{a} \right) + C \). Here, \( a = 2 \).

Thus, we can apply this formula:
\[ I = \frac{1}{2} \left( \sin^{-1} \left( \frac{x}{2} \right) + C \right) \]

### Step 4: Final Result
Therefore, the final result for the integral is:
\[ I = \frac{1}{2} \sin^{-1} \left( \frac{x}{2} \right) + C \]

### Summary
The integral \( \int \frac{dx}{\sqrt{16 - 4x^2}} \) evaluates to:
\[ \frac{1}{2} \sin^{-1} \left( \frac{x}{2} \right) + C \]