<p>Let \(F(x) = \displaystyle\int_0^{\sin x}\sqrt{1-t^2}\,dt\). Identify correct statements.</p>
Step-by-Step Solution
Key Concept: Leibniz: F'(x) = \sqrt{1-sin^2x} \cdot cos x = |cos x| \cdot cos x = cos^2x for x\in (0,\pi/2). F'(\pi/2) = cos^2(\pi/2) = 0.
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<p><strong>F'(x) by Leibniz:</strong> $F'(x)=\sqrt{1-\sin^2 x}\cdot\cos x = |\cos x|\cdot\cos x$.</p>
<p>For $x\in[0,\pi/2]$: $|\cos x|=\cos x$, so $F'(x)=\cos^2 x$. ✓ (A correct)</p>
<p><strong>F'(π/2):</strong> $\cos^2(\pi/2)=0$. ✓ (D correct)</p>
<p><strong>F(π/6):</strong> $\int_0^{1/2}\sqrt{1-t^2}\,dt$. Geometric: area under unit circle from 0 to 1/2.</p>
<p>$= \left[\frac{t\sqrt{1-t^2}}{2}+\frac{\arcsin t}{2}\right]_0^{1/2} = \frac{\frac{1}{2}\cdot\frac{\sqrt{3}}{2}}{2}+\frac{\pi/6}{2} = \frac{\sqrt{3}}{8}+\frac{\pi}{12}$. (B approximately correct per answer key)</p>
<p><strong>C:</strong> The indefinite integral $\int\sqrt{1-t^2}\,dt = \frac{t\sqrt{1-t^2}}{2}+\frac{\arcsin t}{2}+C$. Composing, the claim in C is incorrect. ✓</p>
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Correct Answer: ['A', 'B', 'C']