Question

apply integration by parts twice to evaluate each integral. $$ \int \cos (\ln x) d x $$

Short answer

\( \int \cos(\ln x) dx = \frac{1}{2}(x \cos(\ln x) + x \sin(\ln x)) + C \).

Step-by-step solution

Identify the Functions

We need to use integration by parts, which follows the formula: \( \int u \, dv = uv - \int v \, du \). First, identify \( u \) and \( dv \). Let \( u = \cos(\ln x) \) and \( dv = dx \).

Find du and v

Differentiate \( u \) to find \( du \) and integrate \( dv \) to find \( v \). We have \( du = -\sin(\ln x) \cdot \frac{1}{x} \, dx \) and \( v = x \).

Apply Integration by Parts

Using the integration by parts formula, we substitute \( u \), \( du \), \( v \), and \( dv \) into the formula: \( \int u \, dv = uv - \int v \, du \). We get \( x \cdot \cos(\ln x) - \int x(-\sin(\ln x) \cdot \frac{1}{x}) dx \).

Simplify Inner Integral

Simplify the inner integral: \( \int x(-\sin(\ln x) \cdot \frac{1}{x}) dx = -\int \sin(\ln x) dx \). Now we have the integral \( x \cos(\ln x) + \int \sin(\ln x) dx \).

Second Integration by Parts Preparation

For \( \int \sin(\ln x) dx \), set \( u = \sin(\ln x) \) and \( dv = dx \). Differentiating, \( du = \cos(\ln x) \cdot \frac{1}{x} \, dx \), and integrating, \( v = x \).

Apply Integration by Parts Again

Using integration by parts, \( \int u \, dv = uv - \int v \, du \), we substitute to get \( x \cdot \sin(\ln x) - \int x (\cos(\ln x) \cdot \frac{1}{x}) dx \). Simplifying gives us \( x\sin(\ln x) - \int \cos(\ln x) dx \).

Solve Recurrence

We notice the integral \( \int \cos(\ln x) dx \) is both the original integral and appears again. Substitute known parts from previous steps into the main equation. Let \( I = \int \cos(\ln x) dx \), giving \( I = x \cos(\ln x) + x \sin(\ln x) - I \).

Finalize the Solution

Rearrange the expression to isolate \( I \): \( 2I = x \cos(\ln x) + x \sin(\ln x) \). Solve for \( I \): \( I = \frac{1}{2}(x \cos(\ln x) + x \sin(\ln x)) + C \), where \( C \) is the constant of integration.

Key concepts

Integral Calculus

Integral Calculus is a crucial branch of calculus that focuses on the concepts of integrals and antiderivatives. It enables us to find the total accumulation of quantities, understand the area under curves, and solve problems related to rates of change. The backbone of integral calculus is the concept of the indefinite integral, which represents a family of functions. The process of finding an indefinite integral is referred to as integration.

When we perform integration, we often deal with expressions involving different functions. A crucial formula in integral calculus is the basic form of Integration by Parts, which is a technique used to integrate the product of two functions. It is expressed as:\[ \int u \, dv = uv - \int v \, du \]
This formula is an essential tool especially for functions that are products of polynomial, logarithmic, and trigonometric functions.

• Recognize when an integral cannot be solved by basic antiderivative techniques alone.
• For complex integrals, breaking them into simpler parts often involves using substitution or integration by parts.

Integration by parts is highly beneficial in solving integrals that standard methods may not easily tackle, requiring us to identify parts of the integral that fit the formula above.

Trigonometric Integrals

Trigonometric Integrals involve integrals containing trigonometric functions such as sine, cosine, and tangent. These are frequently encountered in calculus and require specific strategies to integrate.

Understanding the properties and identities of trigonometric functions is crucial. For instance, with the problem \[\int \cos(\ln x) \, dx\], we deal with a composite function involving both logarithmic and trigonometric elements.

• Here, the challenge lies in managing integrals involving oscillatory components like \( \sin(x) \) and \( \cos(x) \).
• Techniques like trigonometric identities and substitution can simplify the integrals involving such functions.

By breaking down the integral using Integration by Parts twice, students can effectively tackle trigonometric integrals. This method ensures that even when faced with complex expressions, a systematic approach can yield a solution.

Calculus Techniques

Various techniques in calculus provide powerful methods for solving integrals that may appear daunting at first glance. One of these is Integration by Parts, used effectively to evaluate the given integral in the problem:
\[ \int \cos(\ln x) \, dx \].

By decomposition into simpler expressions, Integration by Parts helps in solving integrals whose antiderivatives are not immediately obvious. The solution provided for the integral requires using Integration by Parts twice to simplify the problem. Here's a glance at the key steps:

• First, decompose the integral using \( u = \cos(\ln x) \) and \( dv = dx \).
• After simplifying the inner integral, encounter it again and apply Integration by Parts once more.
• Recognizing recurrence in integral expressions necessitates rearranging and solving for the original integral term.

The final solution involves identifying that the original integral reappears and can be solved through algebraic manipulation, ultimately leading to understanding how calculations intersect at various stages in calculus.