To solve the problem, we must find $g'(x)$ where

$g(x) = \int_{1}^{\sin(x)} t^2 \, dt.$

This is a problem that involves the Fundamental Theorem of Calculus and chain rule differentiation.

Step 1: Apply the Fundamental Theorem of Calculus

The derivative of an integral of the form $F(x) = \int_{a(x)}^{b(x)} f(t) \, dt$ is given by: $F'(x) = f(b(x)) b'(x) - f(a(x)) a'(x),$ where $b(x)$ is the upper limit and $a(x)$ is the lower limit.

Here:

• The lower limit $a(x) = 1$ (constant, so $a'(x) = 0$),
• The upper limit $b(x) = \sin(x)$, so $b'(x) = \cos(x)$,
• The integrand $f(t) = t^2$.

Step 2: Differentiate $g(x)$

Using the formula above, we have: $g'(x) = f(\sin(x)) \cdot b'(x) - f(1) \cdot a'(x).$

Substitute $f(t) = t^2$: $g'(x) = (\sin(x))^2 \cdot \cos(x) - 0.$

Thus: $g'(x) = \sin^2(x) \cos(x).$

Step 3: Match with the Choices

The correct answer is: $\text{(d) } \sin^2(x) \cos(x).$

Would you like further explanation or more examples? Here are some related questions to expand your understanding:

1. What is the Fundamental Theorem of Calculus and how does it apply to variable limits?
2. How do we differentiate integrals with both upper and lower limits as functions?
3. Why does the constant lower limit $a(x) = 1$ contribute nothing to the derivative?
4. What happens if the integrand $f(t)$ were more complex, such as $e^{t^2}$?
5. How does the chain rule interact with definite integrals?

Tip: When differentiating an integral with a variable limit, focus on applying the Fundamental Theorem of Calculus step by step to avoid errors!