To find an antiderivative of $b(x) = \sqrt{\cos(7x + 2)}$ using the Second Fundamental Theorem of Calculus and ensure it passes through the point $(12,7)$, we proceed as follows:

1. Antiderivative Construction:
   The Second Fundamental Theorem of Calculus states that if $F(x)$ is defined as: $F(x) = \int_{a}^{x} b(t) \, dt + C,$ then $F'(x) = b(x)$. Here, $b(x) = \sqrt{\cos(7x + 2)}$, and we need to find an antiderivative that passes through the point $(12,7)$.

2. Set Up the Antiderivative:
   We construct $F(x)$ as: $F(x) = \int_{a}^{x} \sqrt{\cos(7t + 2)} \, dt + C.$ We will choose $a = 0$ for simplicity, making the antiderivative: $F(x) = \int_{0}^{x} \sqrt{\cos(7t + 2)} \, dt + C.$

3. Determine the Constant $C$:
   To find $C$, we use the given point $(12, 7)$. This means: $F(12) = 7.$ Thus, we need to solve: $\int_{0}^{12} \sqrt{\cos(7t + 2)} \, dt + C = 7.$ Let $I = \int_{0}^{12} \sqrt{\cos(7t + 2)} \, dt$, then: $I + C = 7 \implies C = 7 - I.$

4. Final Antiderivative Expression:
   The antiderivative is: $F(x) = \int_{0}^{x} \sqrt{\cos(7t + 2)} \, dt + (7 - I),$ where $I$ is the definite integral $\int_{0}^{12} \sqrt{\cos(7t + 2)} \, dt$.

This expression provides an antiderivative that contains the point $(12, 7)$.

Would you like further details or have any questions?

Here are some related questions:

1. How do you calculate definite integrals numerically?
2. What is the significance of choosing different lower limits $a$ in the Second Fundamental Theorem of Calculus?
3. How does the Second Fundamental Theorem of Calculus relate to the First Fundamental Theorem of Calculus?
4. What are the methods to solve integrals involving trigonometric functions?
5. How can you verify if a function is a valid antiderivative?

Tip: When finding antiderivatives, always include the constant of integration $C$, which accounts for all possible solutions.