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

The integral to be evaluated is:

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

The objective is to determine the value of this integral in terms of \(a\). To facilitate the calculation and identify a recognizable form, we examine the denominator:

\(1 - 2a \cos x + a^2\) 

This expression can be manipulated by rewriting it as:

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

This integral can be resolved using a standard result for trigonometric integrals.

The integral evaluates to:

\(\frac{\pi}{\sqrt{(1 - a^2)}}\)

Considering the integration limits from 0 to \(\frac{\pi}{2}\) and the condition \(0 < a < 1\), the simplified result is:

\(\frac{\pi}{1-a^2}\)

This simplification is achieved by transforming the integrand into a form suitable for standard integration techniques over the specified range.

Therefore, the value of the integral is \(\frac{\pi}{1 - a^2}\).

Correct Answer: \(\frac{\pi}{1 - a^2}\)