The value of `int(2dx)/sqrt(1-4x^(2))` is

Options

A. `tan^(-1)(2x)+c`
B. `cos^(-1)(2x)+c`
C. `cot^(-1)(2x)+c`
D. `sin^(-1)(2x)+c`

Answer

To solve the integral \( I = \int \frac{2x \, dx}{\sqrt{1 - 4x^2}} \), we can use a substitution method. Here’s a step-by-step solution:

### Step 1: Substitution
Let us use the substitution:
\[ u = 1 - 4x^2 \]
Then, we differentiate \( u \) with respect to \( x \):
\[ \frac{du}{dx} = -8x \quad \Rightarrow \quad du = -8x \, dx \quad \Rightarrow \quad dx = \frac{du}{-8x} \]

### Step 2: Rewrite the integral
Now we can rewrite the integral in terms of \( u \):
\[ I = \int \frac{2x \cdot \frac{du}{-8x}}{\sqrt{u}} = \int \frac{-2}{8} \frac{du}{\sqrt{u}} = -\frac{1}{4} \int u^{-1/2} \, du \]

### Step 3: Integrate
Now we can integrate:
\[ -\frac{1}{4} \int u^{-1/2} \, du = -\frac{1}{4} \cdot 2u^{1/2} + C = -\frac{1}{2} \sqrt{u} + C \]

### Step 4: Substitute back
Now we substitute back \( u = 1 - 4x^2 \):
\[ I = -\frac{1}{2} \sqrt{1 - 4x^2} + C \]

### Final Result
Thus, the value of the integral is:
\[ I = -\frac{1}{2} \sqrt{1 - 4x^2} + C \]

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