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The Fundamental Theorem of Calculus sits at the heart of integral calculus and bridges the gap between antiderivatives and definite integrals. It is split into two parts: the first part provides a way to evaluate the indefinite integral of a function, while the second part allows for the easy computation of definite integrals.

The theorem states that if a function 'f' is continuous over an interval [a, b], and F is an antiderivative of f in this interval, then the definite integral of f from a to b is given by \( F(b) - F(a) \).

In practice, this means you can calculate the area under the curve of f between a and b by finding the antiderivative F and evaluating it at the endpoints of the interval. Using this theorem reduces the problem of computing a definite integral to finding an antiderivative—a significant simplification.

The process of finding an antiderivative—or the function whose derivative gives the original function—is a pivotal step in calculating definite integrals. For simple functions, there are standard formulas you can apply directly, much like in the exercise with the sine function.

For more complex functions, you might need to use integration techniques such as substitution, integration by parts, or partial fractions. Calculating the antiderivative correctly is crucial; even a small mistake can lead to an incorrect evaluation of the definite integral. Careful manipulation of constants and attention to detail when applying formulas will yield the function that, when differentiated, returns the original function under the integral sign.

One of the basic and often encountered integrals in calculus is that of the sine function. The integral of the sine function \( \sin(ax + b) \) is \( -\frac{1}{a}\cos(ax + b) + C \) where 'a' and 'b' are constants, and 'C' is the constant of integration for indefinite integrals.

When dealing with definite integrals, like in the given exercise, the constant of integration is not needed as it cancels out when computing \( F(b) - F(a) \). For the evaluation process, it's essential to remember to multiply by \( -\frac{1}{a} \) after integrating to compensate for the 'a' value when applying the chain rule in reverse. This straightforward method for integrating the sine function often serves as one of the initial examples of how integration is utilised to solve practical problems in mathematics.