Step-by-step Solution:

Correct Answer: C (-2)

This problem requires the use of the Leibniz integral rule, which is a part of the Fundamental Theorem of Calculus used to find the derivative of an integral where the limits of integration are functions of the variable.

Step 1: The Leibniz Integral Rule

The rule states that if you have an integral of the form $F(x) = \int_{a(x)}^{b(x)} f(t) dt$, its derivative with respect to $x$ is given by:

$$ F'(x) = \frac{d}{dx} \int_{a(x)}^{b(x)} f(t) dt = f(b(x)) \cdot b'(x) - f(a(x)) \cdot a'(x) $$

Step 2: Apply the Rule to the Given Problem

Let's identify the components from the given expression:

• The function being integrated is $f(t) = e^{t^2}$.
• The lower limit is $a(x) = \sin^2 x$.
• The upper limit is $b(x) = 2\sin x$.

Next, we find the derivatives of the limits with respect to $x$:

• $a'(x) = \frac{d}{dx}(\sin^2 x) = 2\sin x \cos x$
• $b'(x) = \frac{d}{dx}(2\sin x) = 2\cos x$

Now, we substitute these into the Leibniz rule formula:

$\frac{d}{dx} \int_{\sin^2 x}^{2\sin x} e^{t^2} dt = f(2\sin x) \cdot (2\cos x) - f(\sin^2 x) \cdot (2\sin x \cos x)$

$= e^{(2\sin x)^2} \cdot (2\cos x) - e^{(\sin^2 x)^2} \cdot (2\sin x \cos x)$

$= e^{4\sin^2 x}(2\cos x) - e^{\sin^4 x}(2\sin x \cos x)$

Step 3: Evaluate the Expression at $x = \pi$

We need to find the value of the derivative at $x = \pi$. We know that:

• $\sin(\pi) = 0$
• $\cos(\pi) = -1$

Substitute these values into the expression for the derivative:

Value $= e^{4(\sin \pi)^2}(2\cos \pi) - e^{(\sin \pi)^4}(2\sin \pi \cos \pi)$

$= e^{4(0)^2}(2(-1)) - e^{(0)^4}(2(0)(-1))$

$= e^0(-2) - e^0(0)$

$= (1)(-2) - (1)(0) = -2$