`int_(0)^(pi) dx/(1-2acos x + a^(2)) =`

Options

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

Answer

To solve the integral

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

we can follow these steps:

### Step 1: Rewrite the Integral
The expression in the denominator can be rewritten using the identity for cosine. We have:

\[ 1 - 2a \cos x + a^2 = (1 - a e^{ix})(1 - a e^{-ix}) \]

This suggests that we can use a substitution involving the tangent half-angle.

### Step 2: Use the Tangent Half-Angle Substitution
Let \( t = \tan\left(\frac{x}{2}\right) \). Then, we have:

\[ \cos x = \frac{1 - t^2}{1 + t^2} \]
and
\[ dx = \frac{2}{1 + t^2} dt \]

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

### Step 3: Substitute in the Integral
Substituting these into the integral gives:

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

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

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

This simplifies to:

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

### Step 5: Rewrite the Integral
Thus, the integral becomes:

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

### Step 6: Factor Out Constants
Let \( A = 1 + a^2 - 2a \) and \( B = 1 + a^2 + 2a \):

\[ I = \frac{2}{B} \int_{0}^{\infty} \frac{dt}{\frac{A}{B} + t^2} \]

### Step 7: Evaluate the Integral
The integral

\[ \int_{0}^{\infty} \frac{dt}{c + t^2} = \frac{\pi}{2\sqrt{c}} \]

for \( c = \frac{A}{B} \) gives:

\[ I = \frac{2}{B} \cdot \frac{\pi}{2\sqrt{\frac{A}{B}}} = \frac{\pi}{\sqrt{AB}} \]

### Step 8: Substitute Back for A and B
Now substituting back:

\[ A = 1 + a^2 - 2a, \quad B = 1 + a^2 + 2a \]

Thus,

\[ AB = (1 + a^2 - 2a)(1 + a^2 + 2a) = (1 + a^2)^2 - (2a)^2 = 1 + 2a^2 + a^4 - 4a^2 = 1 - 2a^2 + a^4 \]

### Final Result
Therefore, we have:

\[ I = \frac{\pi}{\sqrt{(1 - 2a^2 + a^4)}} \]