Problem 46 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 37-48, apply integration by parts twice to evaluate each integral (see Examples 5 and 6). $$ \int \cos (\ln x) d x $$

Step-by-Step Solution

Answer

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

Step 1: Choose Functions for Integration by Parts

Integration by parts is given by the formula \( \int u \, dv = uv - \int v \, du \). We start by setting \( u = \cos(\ln x) \) and \( dv = dx \).

Step 2: Compute Derivatives and Integrals

Differentiate \( u \) and integrate \( dv \):\[- \begin{align*}du &= -\sin(\ln x) \cdot \frac{1}{x} \, dx, \v &= x.\end{align*}\]

Step 3: Apply Integration by Parts First Time

Substituting into the integration by parts formula gives:\[\int \cos(\ln x) \, dx = x \cos(\ln x) - \int x \left( -\sin(\ln x) \cdot \frac{1}{x} \right) dx.\] Simplifying, we have:\[\int \cos(\ln x) \, dx = x \cos(\ln x) + \int \sin(\ln x) \, dx.\]

Step 4: Apply Integration by Parts Second Time

Apply integration by parts to \( \int \sin(\ln x) \, dx \). Choose \( u = \sin(\ln x) \) and \( dv = dx \), so:\[- \begin{align*}du &= \cos(\ln x) \cdot \frac{1}{x} \, dx, \v &= x.\end{align*}\]

Step 5: Calculate Derivatives and Integrals Again

Substitute \( u \) and \( dv \) into the formula:\[\int \sin(\ln x) \, dx = x \sin(\ln x) - \int x \left( \cos(\ln x) \cdot \frac{1}{x} \right) dx.\] This simplifies to:\[\int \sin(\ln x) \, dx = x \sin(\ln x) - \int \cos(\ln x) \, dx.\]

Step 6: Solve the Equation

We arrived at the equation:\[ \int \cos(\ln x) \, dx = x \cos(\ln x) + x \sin(\ln x) - \int \cos(\ln x) \, dx. \]Let \( I = \int \cos(\ln x) \, dx \). Then, substituting, we have:\[ I = x \cos(\ln x) + x \sin(\ln x) - I. \]Solving for \( I \), we get:\[ 2I = x \cos(\ln x) + x \sin(\ln x), \] \[ I = \frac{x}{2} \left( \cos(\ln x) + \sin(\ln x) \right). \]

Step 7: Include Constant of Integration

Finally, include the constant of integration \( C \):\[ \int \cos(\ln x) \, dx = \frac{x}{2} \left( \cos(\ln x) + \sin(\ln x) \right) + C. \]

Key Concepts

Definite Integrals

Definite integrals are a fundamental concept in calculus that allow us to compute the area under a curve within a specific interval. They are represented by an integral sign with limits of integration, which define the interval over which the function is being integrated. The notation for a definite integral of a function \( f(x) \) over the interval \([a, b]\) is given by:\[\int_{a}^{b} f(x) \, dx\]This expression gives the net area between the curve \( f(x) \) and the x-axis from \( x = a \) to \( x = b \). It's important to differentiate between definite and indefinite integrals; while definite integrals compute an actual number (the net area), indefinite integrals represent a family of functions.

• Definite integrals have specified limits of integration.
• They are useful in finding areas, volumes, and many other applications.
• Related to the Fundamental Theorem of Calculus, linking derivatives and integrals.

Understanding how to evaluate indefinite integrals is crucial because definite integrals often rely on these calculations. In the context of the problem at hand, although it's an indefinite integral, grasping definite integrals might help in applications like determining areas when the function \( \cos(\ln x) \) is involved.

Trigonometric Integrals

Trigonometric integrals involve integrating functions that include trigonometric functions such as sine, cosine, tangent, etc. In our exercise, the integral of \( \cos(\ln x) \) is a trigonometric function because it has a cosine component. Solving trigonometric integrals often requires advanced techniques like integration by parts, trigonometric identities, or substitutions, especially when dealing with composite functions like \( \cos(\ln x) \).

• Trigonometric functions are periodic, meaning they repeat values in regular intervals.


• The cosine function relates to the unit circle and baseline in oscillations.


• In trigonometric integrals, familiar formulas and functions like sine, cosine, and their relationships, help simplify processes.

In this problem, integrating \( \cos(\ln x) \) required leveraging integration by parts twice, which is essential when direct integration is not straightforward. It's important to break down composite trigonometric functions, sometimes transforming them using identities before applying integration techniques.

Logarithmic Functions

Logarithmic functions are the inverses of exponential functions and are represented by the function \( \ln x \), denoting the natural logarithm of \( x \). These functions appear in many areas of calculus due to their unique properties of transforming multiplication into addition and supporting constant rates of growth or decay.

• Integration involving logarithms can be challenging, often requiring specific techniques and substitutions.
• The derivative of \( \ln x \) is \( \frac{1}{x} \), a key concept when applying integration by parts.


• Logarithmic functions are useful in solving growth and decay problems, as well as in calculations involving compound interest.

In the given exercise, the presence of \( \ln x \) inside the cosine function makes it necessary to combine the understanding of trigonometric and logarithmic properties to solve the problem effectively. The chain rule also plays a vital role in deriving \( du \) as part of integration by parts.

Calculus Problem Solving

Calculus problem solving involves strategic approaches and techniques to tackle problems like integrals, derivatives, limits, and more. The integration by parts method, used in this problem, is a direct application of a problem-solving strategy in calculus.

• Integration by parts formula is \( \int u \, dv = uv - \int v \, du\), a vital tool for integrating products of functions.
• Identifying \( u \) and \( dv \) correctly is crucial and often depends on simplifying the integrand.


• Problem solving in calculus often requires creativity and breaking down complex problems into more manageable parts.

The solution to the problem combines many concepts: trigonometric identities, the relationship between logarithms and integration, and the perseverance needed to apply integration by parts twice. Each of these steps is necessary to simplify and ultimately solve the integral successfully. This demonstrates how integral calculus extends simple integration to tackle more complex functions.