`int _(0) ^(1//2) dx/((1-2x^(2))sqrt(1-x^(2)))=`

Options

A. `(-1)/2 log (2+sqrt(3))`
B. `(1)/2 log (2+sqrt(3))`
C. `(-1)/4 log (2+sqrt(3))`
D. `(1)/4 log (2+sqrt(3))`

Answer

To solve the integral

\[ I = \int_0^{\frac{1}{2}} \frac{dx}{(1 - 2x^2) \sqrt{1 - x^2}}, \]

we will use the substitution method. Here are the steps:

### Step 1: Substitution
Let \( x = \sin \theta \). Then, we have:

\[ dx = \cos \theta \, d\theta. \]

### Step 2: Change the limits of integration
When \( x = 0 \):
\[ \sin \theta = 0 \implies \theta = 0. \]
When \( x = \frac{1}{2} \):
\[ \sin \theta = \frac{1}{2} \implies \theta = \frac{\pi}{6}. \]

So, the limits change from \( 0 \) to \( \frac{\pi}{6} \).

### Step 3: Rewrite the integral
Substituting \( x \) and \( dx \) into the integral gives:

\[ I = \int_0^{\frac{\pi}{6}} \frac{\cos \theta \, d\theta}{(1 - 2\sin^2 \theta) \sqrt{1 - \sin^2 \theta}}. \]

Since \( \sqrt{1 - \sin^2 \theta} = \cos \theta \), we can simplify the integral:

\[ I = \int_0^{\frac{\pi}{6}} \frac{\cos \theta \, d\theta}{(1 - 2\sin^2 \theta) \cos \theta} = \int_0^{\frac{\pi}{6}} \frac{d\theta}{1 - 2\sin^2 \theta}. \]

### Step 4: Use the identity for \( 1 - 2\sin^2 \theta \)
Recall that \( 1 - 2\sin^2 \theta = \cos 2\theta \). Thus, we can rewrite the integral as:

\[ I = \int_0^{\frac{\pi}{6}} \frac{d\theta}{\cos 2\theta}. \]

### Step 5: Change to secant
This can be expressed as:

\[ I = \int_0^{\frac{\pi}{6}} \sec 2\theta \, d\theta. \]

### Step 6: Integrate
The integral of \( \sec kx \) is given by:

\[ \int \sec kx \, dx = \frac{1}{k} \ln | \sec kx + \tan kx | + C. \]

In our case, \( k = 2 \):

\[ I = \frac{1}{2} \ln | \sec 2\theta + \tan 2\theta | \bigg|_0^{\frac{\pi}{6}}. \]

### Step 7: Evaluate the limits
Now we evaluate at the limits:

1. At \( \theta = \frac{\pi}{6} \):
- \( \sec \frac{\pi}{3} = 2 \)
- \( \tan \frac{\pi}{3} = \sqrt{3} \)

Therefore,

\[ \sec 2\left(\frac{\pi}{6}\right) + \tan 2\left(\frac{\pi}{6}\right) = 2 + \sqrt{3}. \]

2. At \( \theta = 0 \):
- \( \sec 0 = 1 \)
- \( \tan 0 = 0 \)

Therefore,

\[ \sec 0 + \tan 0 = 1. \]

### Step 8: Substitute back into the integral
Putting it all together:

\[ I = \frac{1}{2} \left( \ln(2 + \sqrt{3}) - \ln(1) \right) = \frac{1}{2} \ln(2 + \sqrt{3}). \]

### Final Answer
Thus, the final answer is:

\[ I = \frac{1}{2} \ln(2 + \sqrt{3}). \]