Let $f(x) = \int_{x^2/4}^{x^2} \frac{\sin x}{1+\cos^2 \sqrt{t}} dt$ then
Step-by-Step Solution
Key Concept: Apply Leibniz integral rule: $\frac{d}{dx}\int_{a(x)}^{b(x)} g(t)dt = g(b(x))b'(x) - g(a(x))a'(x)$, being careful with absolute values in the argument.
Using Leibniz rule for differentiation under the integral sign: $f'(x) = \frac{\sin x}{1+\cos^2 \sqrt{x^2}} \cdot 2x - \frac{\sin x}{1+\cos^2 \sqrt{x^2/4}} \cdot \frac{x}{2}$. Since $\sqrt{x^2} = |x|$ and $\sqrt{x^2/4} = |x|/2$, we get $f'(x) = \frac{\sin x}{1+\cos^2|x|}(2x - \frac{x}{2}) = \frac{\sin x}{1+\cos^2|x|} \cdot \frac{3x}{2}$. At $x = \pi/2 > 0$: $f'(\pi/2) = \frac{1}{1+\cos^2(\pi/2)} \cdot \frac{3\pi}{4} = \frac{3\pi}{4}$ (checking options 1,2). At $x = 3\pi/2 > 0$: $f'(3\pi/2) = \frac{-1}{1+\cos^2(3\pi/2)} \cdot \frac{9\pi}{4} = -\frac{9\pi}{4}$ (option 3 needs verification). Option 4 involves integrating $f'(\pi)$ which equals $\frac{\sin\pi}{1+\cos^2\pi} \cdot \frac{3\pi}{2} = 0$, matching the integral evaluation.
Correct Answer: 1,2,3,4