91Ó°ÊÓ

The substitution method is a very useful technique for evaluating integrals, especially when the integral involves a composite function. It simplifies the integration process by making a "change of variables".

In this exercise, we are dealing with the integral \( \int \sin(\ln x) \ dx \). The substitution method helps us transform this integral into a simpler form.

**Steps in Using Substitution:**

• Identify a substitution: Choose a new variable, usually \( u \), which simplifies the expression. Here, \( u = \ln x \).
• Express \( dx \) in terms of \( du \): From \( u = \ln x \), differentiate to get \( du = \frac{1}{x} \, dx \), rearranging gives \( dx = x \, du \).
• Substitute and simplify: Replace all terms in the integral with their expressions in terms of \( u \). For \( x = e^u \), \( dx = e^u \, du \).

After substitution, the integral becomes \( \int \sin(u) \, e^u \, du \). This transformation helps us handle the integral more easily through other techniques, like integration by parts.

Integration by parts originates from the product rule for differentiation. It's very effective when the integral is a product of two functions, as seen in this exercise.

Here, we have already transformed our integral \( \int \sin(u) \, e^u \, du \) using substitution. Now, we apply the integration by parts formula:

\[ \int v \, dw = vw - \int w \, dv \]

Here's how to apply this formula:

• Choose \( v \) and \( dw \) wisely, such that their derivatives (\( dv \) and \( w \)) simplify the problem. For example, in our solution, \( v = e^u \) and \( dw = \sin(u) \, du \).
• Find \( dv \): Differentiate \( v \), yielding \( dv = e^u \, du \).
• Find \( w \): Integrate \( dw \), leading to \( w = -\cos(u) \).
• Apply the formula: Substituting these into the formula, we get \( -e^u \cos(u) - \int e^u (-\cos(u)) \, du \).

This method might need to be applied multiple times, as shown when the remaining integral also required integration by parts. Through persistence, you rearrange components and solve our integral.

Understanding definite and indefinite integrals is key for solving a wide range of calculus problems. In this scenario, we solved an indefinite integral.

An **indefinite integral** gives a family of functions, characterized by a constant of integration \( C \). It is of the form:

\[ \int f(x) \, dx = F(x) + C \]

Here, \( F(x) \) is an antiderivative of \( f(x) \). This means that when taking the derivative of \( F(x) \), we arrive back at \( f(x) \).

In our problem, we concluded with:

\[ \int \sin(\ln x) \, dx = \frac{1}{2} x (\sin(\ln x) - \cos(\ln x)) + C \]

The \( + C \) signifies an indefinite integral, indicating all functions that differ only by a constant value.

In contrast, a **definite integral** calculates the area under a curve between two specified points. It results in a numerical value rather than a function. Grasping the difference between these two types is crucial for correctly understanding solutions and applying techniques like substitution and integration by parts across different contexts.