If `0 lt a lt 1`, then `int_(-1)^(1) (1)/(sqrt(1-2ax+a^(2)))dx` is equal to

Options

A. 1
B. `2pi`
C. `pi//2`
D. `3pi//2`

Answer

To solve the integral

\[ I = \int_{-1}^{1} \frac{1}{\sqrt{1 - 2ax + a^2}} \, dx \]

where \( 0 < a < 1 \), we can follow these steps:

### Step 1: Simplify the expression under the square root
We start with the expression inside the square root:

\[ 1 - 2ax + a^2 \]

This can be rewritten as:

\[ (1 - ax)^2 \]

### Step 2: Substitute
Let’s make the substitution:

\[ T^2 = 1 - 2ax + a^2 \]

Differentiating both sides gives:

\[ dT = -2a \, dx \quad \Rightarrow \quad dx = \frac{-dT}{2a} \]

### Step 3: Change the limits of integration
Now we need to change the limits of integration based on our substitution.

1. When \( x = -1 \):

\[ T^2 = 1 - 2a(-1) + a^2 = 1 + 2a + a^2 = (1 + a)^2 \quad \Rightarrow \quad T = 1 + a \]

2. When \( x = 1 \):

\[ T^2 = 1 - 2a(1) + a^2 = 1 - 2a + a^2 = (1 - a)^2 \quad \Rightarrow \quad T = 1 - a \]

### Step 4: Substitute into the integral
Now substituting \( dx \) and changing the limits:

\[ I = \int_{1 + a}^{1 - a} \frac{1}{\sqrt{(1 - ax)^2}} \cdot \frac{-dT}{2a} \]

Since \( \sqrt{(1 - ax)^2} = |1 - ax| \), and for \( 0 < a < 1 \) and \( x \) in \([-1, 1]\), \( 1 - ax \) is always positive. Thus, we can drop the absolute value:

\[ I = \int_{1 + a}^{1 - a} \frac{1}{1 - ax} \cdot \frac{-dT}{2a} \]

### Step 5: Simplify the integral
Reversing the limits gives:

\[ I = \frac{1}{2a} \int_{1 - a}^{1 + a} \frac{1}{T} \, dT \]

### Step 6: Evaluate the integral
The integral of \( \frac{1}{T} \) is:

\[ \int \frac{1}{T} \, dT = \ln |T| \]

Thus, we have:

\[ I = \frac{1}{2a} \left[ \ln |T| \right]_{1 - a}^{1 + a} \]

Calculating this gives:

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

### Final Result
Thus, the value of the integral is:

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