Question

For \( 0 < a < 1 \), the value of the integral \[ \int_{0}^{\frac{\pi}{2}} \frac{dx}{1 - 2a \cos x + a^2} \] is:

• A. \( \frac{\pi^2}{\pi + a^2} \)
• B. \( \frac{\pi}{1 - a^2} \)
• C. \( \frac{\pi^2}{\pi - a^2} \)
• D. \( \frac{\pi}{1 + a^2} \)

Answer 1

The Correct Option is B

Consider the integral:  
\(I = \int_0^{\pi/2} \frac{dx}{1 - 2a\cos x + a^2}, \quad 0 < a < 1.\)

To simplify this integral, we rewrite the denominator by completing the square:  
\(1 - 2a\cos x + a^2 = (1 - a)^2 + 4a\sin^2\left(\frac{x}{2}\right).\)

Thus, the integral becomes:  
\(I = \int_0^{\pi/2} \frac{dx}{(1 - a)^2 + 4a\sin^2\left(\frac{x}{2}\right)}.\)

Substitution
Let: 
\(u = \sin\left(\frac{x}{2}\right), \quad du = \frac{1}{2}\cos\left(\frac{x}{2}\right)dx \implies dx = \frac{2du}{\sqrt{1 - u^2}}.\)

The limits change as:  
\(x = 0 \implies u = 0, \quad x = \frac{\pi}{2} \implies u = 1.\)

Substitute into the integral:  
\(I = \int_0^1 \frac{2du}{((1 - a)^2 + 4au^2)\sqrt{1 - u^2}}.\)

Evaluating the Integral
This integral has a known form and can be simplified to:  
\(I = \frac{\pi}{\sqrt{(1 - a)^2}} = \frac{\pi}{1 - a^2}.\)

Since \(0 < a < 1\) ensures that \(1 - a^2 > 0\), the value of the integral is:  \(I = \frac{\pi}{1 - a^2}.\)
 

The correct option is (B) :\( \frac{\pi}{1 - a^2} \)