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.
A body of mass 1000 kg is moving horizontally with a velocity of 6 m/s. If 200 kg extra mass is added, the final velocity (in m/s) is: