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

Provided:

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

To solve this integral:

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

The integral transforms to:

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

Step 1: Denominator Simplification Using trigonometric identities:

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

Step 2: Substitution and Integration The integral is now:

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

Step 3: Integral Evaluation This integral, when evaluated using standard trigonometric identities and limits, results in a closed-form expression for \( f(t) \).

Step 4: Second Integral Calculation Consider the integral:

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

Substituting the derived \( f(t) \) and simplifying yields:

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

The final result is 1.