91Ó°ÊÓ

The substitution method is a powerful technique that can simplify many integral problems by transforming the variables. In our problem, we aim to convert the integral \( \int_{1}^{3} \frac{\sin 2x}{x} \, dx \) into a more manageable form.
First, we choose a substitution that will simplify the expression. Here, we use \( u = 2x \). This substitution directly impacts the differential \( dx \); specifically, it changes to \( dx = \frac{du}{2} \). When we substitute for \(x\), we also adjust the limits of integration. For example:

• When \( x = 1 \), \( u = 2 \, (2 \times 1) \)
• When \( x = 3 \), \( u = 6 \, (2 \times 3) \)

These changes align the function with the format of an antiderivative for \( f(x) = \frac{\sin x}{x} \). As you perform the substitution and transformation of limits, the integral transforms, allowing you to express it in terms of the variable \( u \), which can be simpler to evaluate.

Trigonometric functions often appear in integrals, especially involving sine and cosine functions. In integrals like \( \int \frac{\sin u}{u} \, du \), the trigonometric component \( \sin u \) is divided by \( u \, (the argument) \).
Integrating trigonometric functions can sometimes be challenging because they do not always lead directly to elementary functions. The function \( \frac{\sin x}{x} \), known as the sinc function, has a known antiderivative, but it is typically expressed in terms of another function or a given primitive, such as \( F(x) \).
When a definite integral of a trigonometric function does not yield to straightforward antiderivatives, substitution or special functions like \( F(x) \) defined explicitly for \( \frac{\sin x}{x} \), provide alternative ways to solve the integral. This was helpful in expressing \( \int_{2}^{6} \frac{\sin u}{u} du \) in terms of \( F \), bypassing direct computation.

Definite integrals are used to find the area under a curve between two specific limits. In this problem, we were tasked with evaluating \( \int_{1}^{3} \frac{\sin 2x}{x} \, dx \). After applying the substitution method, we ended with a function defined in terms of an antiderivative, \( F(x) \).
Definite integrals differ from indefinite integrals because they provide a specific value instead of a family of functions. They incorporate specific limits of integration (like from 2 to 6 here).
When expressing a definite integral like \( \int_{2}^{6} \frac{\sin u}{u} du \) in terms of a function \( F(x) \), we use the fundamental theorem of calculus, which states:

• The definite integral of a function from \( a \) to \( b \) can be found by evaluating an antiderivative \( F(x) \) at these endpoints \( b \) and \( a \), and taking the difference: \( F(b) - F(a) \).

Using this idea, we determined the definite integral as \( F(6) - F(2) \), providing an exact numerical result when \( F(x) \) is known. This shows the power of working with antiderivatives in solving definite integrals efficiently.