Question

Find the integral by using the simplest method. Not all problems require integration by parts. $$ \int \cos (\ln x) d x $$

Short answer

The integral is \( \frac{1}{2} x (\sin(\ln x) + \cos(\ln x)) + C \).

Step-by-step solution

Identify Suitable Substitution

To solve the integral \( \int \cos(\ln x) \, dx \), we will use the substitution method. Set \( u = \ln x \). Then the differential is \( du = \frac{1}{x} \, dx \), which means \( dx = x \, du \). Since \( x = e^u \), we can write \( dx = e^u \, du \). This simplifies the integral to an expression in terms of \( u \).

Substitute and Transform the Integral

Substitute \( u = \ln x \) into the integral to get \( \int \cos(u) \cdot e^u \, du \). This transformation helps us consider the original problem in terms of \( u \), a potentially simpler variable.

Use Integration by Parts

Apply integration by parts to the integral \( \int e^u \cos(u) \, du \). Let \( v = e^u \) and \( dw = \cos(u) \, du \). This gives \( dv = e^u \, du \) and \( w = \sin(u) \). The integration by parts formula \( \int v \, dw = vw - \int w \, dv \) becomes:\[ \int e^u \cos(u) \, du = e^u \sin(u) - \int \sin(u) \cdot e^u \, du \].

Solve the New Integral with Integration by Parts Again

We now solve \( \int e^u \sin(u) \, du \) using integration by parts again. Let \( v = e^u \) and \( dw = \sin(u) \, du \). Then, \( dv = e^u \, du \) and \( w = -\cos(u) \). Substitute these into the integration by parts formula:\[ \int e^u \sin(u) \, du = -e^u \cos(u) + \int \cos(u) \cdot e^u \, du \].

Equate and Solve for the Original Integral

Notice that the new integral is the same as the original integral. Thus:\[ \int e^u \cos(u) \, du = e^u \sin(u) + e^u \cos(u) - \int e^u \cos(u) \cdot du \].Add \( \int e^u \cos(u) \, du \) to both sides to isolate the integral:\[ 2 \int e^u \cos(u) \, du = e^u \sin(u) + e^u \cos(u) \].Divide by 2 to solve for the integral:\[ \int e^u \cos(u) \, du = \frac{1}{2}[e^u(\sin(u) + \cos(u))] + C \].

Substitute Back to Original Variable

Substitute back \( u = \ln x \) and \( e^u = x \):\[ \int \cos(\ln x) \, dx = \frac{1}{2} x (\sin(\ln x) + \cos(\ln x)) + C \].

Key concepts

Substitution Method

The substitution method is a powerful technique used in calculus to simplify the process of integration. By changing the variable of integration, the integral becomes easier to solve. In this case, for the integral \( \int \cos(\ln x) \, dx \), we set \( u = \ln x \). This substitution converts the complex terms into a manageable form. The differential \( dx \) changes to \( e^u \, du \), making the integral \( \int \cos(u) \cdot e^u \, du \). This transforms a complicated problem into one that resembles a standard form, allowing other integration techniques, such as integration by parts, to be applied more effectively.

• Recognize when to use substitution: look for composite functions like \( \cos(\ln x) \).
• Perform substitution to simplify complex integrals into a single-variable format.
• Ensure to change all parts of the integral to terms of the new variable.

Definite Integrals

Definite integrals represent the area under a curve between two specified limits. When faced with a problem, they provide a solution with precise numerical values. While our exercise involves an indefinite integral, understanding definite integrals can help in situations where bounds are given. If we had definite integrals, we would calculate numerical results rather than leaving an arbitrary constant \( C \) as part of the solution.

• Definite integrals have limits and provide specific numerical results.
• They are useful for finding areas, accumulated quantities, or totals.
• In real-life scenarios, definite integrals calculate exact values between two points.

Indefinite Integrals

In contrast, indefinite integrals represent a family of functions and do not have limits of integration. The aim is to find the antiderivative of a function. The solution includes a constant of integration, \( C \), because antiderivatives are not unique. In our problem \( \int \cos(\ln x) \, dx \), the goal is to find a function whose derivative gives back \( \cos(\ln x) \). This indefinite integral process involves symbolic manipulation rather than numerical calculus, keeping the expression general for any upper and lower limits.

• Indefinite integrals are represented without upper and lower limits.
• They describe a set of functions and include a constant \( C \).
• The process finds a function whose differentiation returns the original integrand.

Exponential Functions

Exponential functions are functions where the variable is in the exponent, typically in the form \( e^x \). These functions often appear in calculus, especially in integration tasks, because of their unique properties related to growth, decay, and transformation skills they require. Here, in our integral \( \int \cos(\ln x) \, dx \), exponential functions manifest when substituting \( u = \ln x \) leading to \( x = e^u \). Understanding exponential functions is crucial as they help transition the integral from a composite function involving logarithms into a more tractable format.

• Exponential functions exhibit continuous growth or decay.
• They are notable in calculus due to their interchangeable nature with derivatives and integrals.
• Transformation skills are essential when exponential functions are involved, as seen with equations like \( x = e^u \).