The integral `int_(0)^(pi) (1)/(a^(2)-2 a cos x+1)dx (a lt 1)` is

Options

A. `(pi)/(1-a^(2))`
B. `(pi)/(a^(2)-1)`
C. `(2pi)/(a^(2)-1)`
D. `(3pi)/(4)`

Answer

To solve the integral

\[ I = \int_{0}^{\pi} \frac{1}{a^2 - 2a \cos x + 1} \, dx \]

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

### Step 1: Rewrite the Denominator
The expression in the denominator can be rewritten using the identity for cosine:

\[ a^2 - 2a \cos x + 1 = (a - \cos x)^2 + \sin^2 x \]

This helps to express the integral in a more manageable form.

### Step 2: Change of Variables
Next, we can use the substitution \( t = \tan\left(\frac{x}{2}\right) \). The differential \( dx \) can be expressed as:

\[ dx = \frac{2}{1+t^2} \, dt \]

The limits of integration change as follows:
- When \( x = 0 \), \( t = 0 \)
- When \( x = \pi \), \( t \to \infty \)

### Step 3: Substitute in the Integral
Now, substituting \( x \) and \( dx \) into the integral gives:

\[ I = \int_{0}^{\infty} \frac{2}{(a - \frac{1-t^2}{1+t^2})^2 + (1+t^2)^2} \, dt \]

### Step 4: Simplify the Denominator
Now, simplify the denominator:

\[ a - \frac{1-t^2}{1+t^2} = \frac{a(1+t^2) - (1-t^2)}{1+t^2} = \frac{(a+1)t^2 + (a-1)}{1+t^2} \]

Thus, the integral becomes:

\[ I = \int_{0}^{\infty} \frac{2(1+t^2)}{((a+1)t^2 + (a-1))^2 + (1+t^2)^2} \, dt \]

### Step 5: Evaluate the Integral
This integral can be evaluated using the formula for integrals of the form:

\[ \int_{0}^{\infty} \frac{1}{a^2 + b^2 x^2} \, dx = \frac{\pi}{2ab} \]

By identifying appropriate values for \( a \) and \( b \) in our integral, we can find the solution.

### Final Result
After evaluating the integral, we find that:

\[ I = \frac{\pi}{a^2 - 1} \]