Question

Evaluate \(\int \cos (\ln x) d x\) two different ways: a. Use tables after first using the substitution \(u=\ln x\) b. Use integration by parts twice to verify your answer to part (a).

Short answer

Question: Evaluate the integral \(\int \cos(\ln x) d x\) in two different ways: (a) using the substitution \(u = \ln x\) and (b) using integration by parts twice. Verify that the results are equivalent. Answer: Using substitution, the result is \(\sin(\ln x) + C\). Using integration by parts twice, the result is \(\frac{1}{2}x (\cos(\ln x) + \sin(\ln x)) + C'\). These results are equivalent up to a constant.

Step-by-step solution

Part (a) - Use substitution \(u = \ln x\)

First, perform the substitution \(u = \ln x\). To do this, we need to find the corresponding \(d u\). Differentiate the substitution with respect to \(x\): $$\frac{d u}{d x} = \frac{d}{d x} (\ln x) = \frac{1}{x}.$$ Now, multiply both sides by \(d x\) to find \(d u\): $$d u = \frac{d x}{x}.$$ Now substitute \(u\) and \(d u\) into the integral and solve for the new integral: $$\int \cos(\ln x) d x = \int \cos(u) \frac{d x}{x} = \int \cos(u) d u.$$ The integral now becomes a standard integral of the cosine function: $$\int \cos(u) d u = \sin(u) + C.$$ Substitute back the original variable \(x\) by using the relationship \(u = \ln x\): $$\sin(\ln x) + C = \sin(\ln x) + C.$$ This is the final answer for part (a).

Part (b) - Use integration by parts twice

Recall the integration by parts formula: $$\int u d v = u v - \int v d u.$$ Now, choose the function to be differentiated and the one to be integrated. Let \(u = \cos(\ln x)\) and \(d v = d x\). Differentiate \(u\) with respect to \(x\): $$d u = \frac{d}{d x} (\cos(\ln x)) = -\sin(\ln x) \cdot \frac{1}{x} d x = -\frac{\sin(\ln x)}{x} d x.$$ Integrate \(d v\) with respect to \(x\): $$v = \int d x = x.$$ Now, apply the integration by parts formula: $$\int \cos(\ln x) d x = x \cdot \cos(\ln x) - \int x \cdot -\frac{\sin(\ln x)}{x} d x.$$ Simplify the equation: $$\int \cos(\ln x) d x = x \cdot \cos(\ln x) + \int \sin(\ln x) d x.$$ Now, apply integration by parts once more. Choose \(u' = \sin(\ln x)\) and \(d v' = d x\). Differentiate \(u'\) with respect to \(x\): $$d u' = \frac{d}{d x} (\sin(\ln x)) = \cos(\ln x) \cdot \frac{1}{x} d x = \frac{\cos(\ln x)}{x} d x.$$ Integrate \(d v'\) with respect to \(x\): $$v' = \int d x = x.$$ Apply the integration by parts formula again: $$\int \sin(\ln x) d x = x \cdot \sin(\ln x) - \int x \cdot \frac{\cos(\ln x)}{x} d x.$$ Simplify the equation: $$\int \sin(\ln x) d x = x \cdot \sin(\ln x) - \int \cos(\ln x) d x.$$ Now, substitute this result back into the first equation obtained using integration by parts: $$\int \cos(\ln x) d x = x \cdot \cos(\ln x) + x \cdot \sin(\ln x) - \int \cos(\ln x) d x.$$ Rearrange the equation to solve for the integral: $$2 \int \cos(\ln x) d x = x \cdot \cos(\ln x) + x \cdot \sin(\ln x).$$ Divide both sides by 2: $$\int \cos(\ln x) d x = \frac{1}{2}x (\cos(\ln x) + \sin(\ln x)) + C'.$$ Comparing the results of part (a) and part (b), we can verify that they are equivalent (up to a constant \(C'\)): $$\sin(\ln x) + C = \frac{1}{2}x (\cos(\ln x) + \sin(\ln x)) + C'.$$

Key concepts

Substitution Method

The substitution method is an effective tool for evaluating integrals, especially when dealing with compositions of functions. The main idea is to transform a complex integral into a simpler one by substituting a part of the integrand with a new variable. In this case, we used the substitution \(u = \ln x\), which simplifies the integral by reducing it to a basic trigonometric form.
Here's how it works:

• Identify a portion of the integrand that can be replaced. In our example, we chose \( \ln x \).
• Define \( u = \ln x \) and compute \( du \) by differentiating: \( du = \frac{dx}{x} \).
• Substitute \( u \) and \( du \) into the integral, converting it to \( \int \cos(u) \, du \).
• This simpler integral can be solved as \( \sin(u) + C \).
• Finally, revert back to the original variable using \( u = \ln x \).

This method is particularly useful when your integral includes compositions that match a known pattern.

Integration by Parts

Integration by parts is another powerful technique to evaluate integrals. It is modeled after the product rule for differentiation. The goal is to transform an integral of a product of functions into one that is easier to evaluate. The basic formula to keep in mind is:\[ \int u \, dv = u \cdot v - \int v \, du \] In this exercise, integration by parts was used twice to resolve the integral of \( \cos(\ln x) \). Here's a step-by-step breakdown:

• First, let \( u = \cos(\ln x) \) and \( dv = dx \). Compute \( du \) and \( v \):
  • \( du = -\frac{\sin(\ln x)}{x} dx \)
  • \( v = x \)
• Apply the integration by parts formula: \( \int \cos(\ln x) \, dx = x \cdot \cos(\ln x) - \int x \cdot -\frac{\sin(\ln x)}{x} \, dx \).
• Then, apply integration by parts again, this time for \( \int \sin(\ln x) \, dx \).
• After simplifying, bring back the terms together to solve the original integral.
• Compare the expression obtained with the result from using substitution.

Using integration by parts requires careful selection of \( u \) and \( dv \) but can simplify integrals that are products of functions.

Definite and Indefinite Integrals

Integrals are fundamental concepts in calculus. They come in two types: definite and indefinite.

• Indefinite Integrals: These represent a family of functions and include a constant of integration, \( C \). Since integration is the inverse of differentiation, indefinite integrals generalize the function's antiderivative. For example, \( \int \cos(\ln x) \, dx = \sin(\ln x) + C \).
• Definite Integrals: These are used to calculate the area under a curve between two points. They're expressed as \( \int_{a}^{b} f(x) \, dx \), providing a specific numeric value. The limits \( a \) and \( b \) replace any appearance of the constant \( C \), giving a precise answer.

Indefinite integrals set the stage for solving problems while definite integrals are often used to find specific values or areas. Knowing when and how to use each is key in integration techniques.