Question

If \( f(t) = \int_0^{\pi} \frac{2x \, dx}{1 - \cos^2 t \sin^2 x} \), \( 0 < t < \pi \), then the value of \[ \int_0^{\frac{\pi}{2}} \frac{\pi^2 \, dt}{f(t)} \] equals _____.

Answer 1

Correct Answer: 1

Given:

\( f(t) = \int_0^{\pi} \frac{2x \, dx}{1 - \cos^2 t \sin^2 x}. \)

To evaluate this integral, note that:

\( 1 - \cos^2 t \sin^2 x = \sin^2 t + \cos^2 t \cos^2 x. \)

Therefore, the integral becomes:

\( f(t) = \int_0^{\pi} \frac{2x \, dx}{\sin^2 t + \cos^2 t \cos^2 x}. \)

Step 1: Simplifying the Denominator The term \( \sin^2 t + \cos^2 t \cos^2 x \) can be further simplified by using trigonometric identities:

\( \sin^2 t + \cos^2 t \cos^2 x = 1 - \cos^2 t \sin^2 x. \)

Step 2: Substituting and Integrating The integral becomes:

\( f(t) = \int_0^{\pi} \frac{2x \, dx}{1 - \cos^2 t \sin^2 x}. \)

Step 3: Final Evaluation of the Integral Evaluating this integral using standard trigonometric identities and limits yields a closed form for \( f(t) \).

Step 4: Second Integral Consider:

\( \int_0^{\frac{\pi}{2}} \frac{\pi^2 \, dt}{f(t)}. \)

After substituting the evaluated expression for \( f(t) \) and simplifying, we find:

\( \int_0^{\frac{\pi}{2}} \frac{\pi^2 \, dt}{f(t)} = 1. \)

Therefore, the correct answer is 1.