Answer is: option3
\( 1 \)Solution:
If:
\( F(x) = \int_a^{g(x)} f(t)\,dt \),
then:
\( F'(x) = f(g(x)) \cdot g'(x) \)
In our case:
- \( f(t) = \frac{1}{\sqrt{1 - t^2}} \)
- \( g(x) = \sin x \), so \( g'(x) = \cos x \)
Apply the rule:
\( F'(x) = \frac{1}{\sqrt{1 - (\sin x)^2}} \cdot \cos x \)
\( 1 - \sin^2 x = \cos^2 x \Rightarrow \sqrt{1 - \sin^2 x} = |\cos x| \)
But since \( x \in \left( -\frac{\pi}{2}, \frac{\pi}{2} \right) \), we know \( \cos x > 0 \), so:
\( F'(x) = \frac{1}{\cos x} \cdot \cos x = 1 \)
Hence Option C