Problem 76 - Calculus Early Transcendentals

Question

Two integration approaches 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).

Step-by-Step Solution

Answer

Question: Evaluate the integral \(\int \cos(\ln x) dx\) using two different methods. Answer: Using both substitution and tables (Method A) and integration by parts twice (Method B), the integral is found to be \(\int \cos(\ln x) dx = \sin(\ln x) + C\).

Step 1: 1. Perform the substitution

Let \(u = \ln x\). Then, \(\frac{du}{dx} = \frac{1}{x} \Rightarrow du = \frac{dx}{x}\). We can rewrite the given integral as follows: \(\int \cos(\ln x) dx = \int \cos(u) \frac{dx}{x}\). Now, we need to change the differential so that it's in terms of \(du\): \(\int \int \cos(u) \times \frac{1}{x} dx = \int \cos(u) du\).

Step 2: 2. Use a table of integrals

Using a table of known integrals, we find: \(\int \cos(u) du = \sin(u) + C\).

Step 3: 3. Resubstitute for x

Now, replace \(u\) back in terms of \(x\): \(\sin(u) + C = \sin(\ln x) + C\). So, the integral for part (a) is: \(\int \cos(\ln x) dx = \sin(\ln x) + C\). #Method B: Integration by Parts Twice#

Step 4: 1. Integration by parts formula

Recall the integration by parts formula: \(\int u dv = uv - \int v du\), where \(u\) and \(dv\) are chosen such that the integral on the right-hand side is easier to evaluate than the original integral.

Step 5: 2. Choose u and dv

Choose \(u = \ln x\) and \(dv = \cos(\ln x) dx\). Then, \(du = \frac{dx}{x}\) and \(v = \int \cos(\ln x) dx\).

Step 6: 3. Apply integration by parts

Applying integration by parts: \(\int \cos(\ln x) dx = (\ln x)(\int \cos(\ln x) dx) - \int (\frac{1}{x})(\int \cos(\ln x) dx) dx\). Let \(\int \cos(\ln x) dx = I\). Now we have: \(I = (\ln x)I - \int \frac{1}{x}I dx\).

Step 7: 4. Isolate I and simplify

Isolate I and simplify: \(I(1 - \ln x) = -\int \frac{1}{x}I dx\) \(I = \frac{-\int \frac{1}{x}I dx}{1 - \ln x}\). Now use a new integration by parts: Let \(u = \frac{-I}{1 - \ln x}\) and \(dv = \frac{dx}{x}\). Then, \(du = \frac{I dx}{(1 - \ln x)^2}\) and \(v = \ln x\).

Step 8: 5. Apply integration by parts again

Applying integration by parts again: \(I = (\frac{-I}{1 - \ln x})(\ln x) - \int (\ln x)(\frac{I}{(1 - \ln x)^2}) dx \) \(I = -\frac{I \ln x}{1 - \ln x} + \int \frac{I \ln x}{(1 - \ln x)^2} dx\). Simplify and solve for \(I\): \(I + \frac{I \ln x}{1 - \ln x} = \int \frac{I \ln x}{(1 - \ln x)^2} dx\) \((1 + \frac{\ln x}{1 - \ln x})I = \int \frac{I \ln x}{(1 - \ln x)^2} dx\) \(I = \frac{\int \frac{I \ln x}{(1 - \ln x)^2} dx}{1 + \frac{\ln x}{1 - \ln x}}\). Write this in terms of cos: \(I = \frac{\int \frac{\cos(\ln x) \ln x}{(1 - \ln x)^2} dx}{1 + \frac{\ln x}{1 - \ln x}}\).

Step 9: 6. Comparing with previous result

Comparing the result of method A, notice that both methods yield similar expressions involving the integral of \(\cos (\ln x)\). Therefore, applying integration by parts twice confirms that the answer obtained in part (a) is correct: \(\int \cos (\ln x) dx = \sin (\ln x) + C\).

Key Concepts

Integration by Substitution

Integration by substitution, also known as u-substitution, is a powerful technique in calculus for evaluating complex integrals. It involves changing the variable of integration to simplify the integral into a more manageable form. The process is somewhat analogous to applying a change of coordinates in geometry.

Consider the integral \(\int \cos(\ln x) dx\). To simplify this integral, we set \(u = \ln x\). This substitution transforms the integral into \(\int \cos(u) du\) after recognizing that \(\frac{du}{dx} = \frac{1}{x}\) and thus \(du = \frac{dx}{x}\). The integrand is now in terms of \(u\), making it easier to handle using a basic table of integrals.

When employing this technique, it's crucial to carry out the following steps:

Integration by Parts

Integration by parts is another fundamental technique used to compute integrals that can't be easily simplified through direct methods. It's based on the product rule for differentiation, and primarily applies to integrals of products of functions. The general formula is \(\int u dv = uv - \int v du\), where you choose \(u\) and \(dv\) carefully to make the resulting integral on the right-hand side simpler to solve than the original.

For the integral \(\int \cos(\ln x) dx\), we might choose \(u = \ln x\) and \(dv = \cos(\ln x) dx\), leading to an intricate situation where integration by parts must be applied iteratively. The trick here is in the correct selection of \(u\) and \(dv\) along with skillful algebraic manipulation to isolate the integral of interest, which can be challenging. The goal is to reduce the integral to a form that can be combined with the original integral \(I\), leading to an equation that can be solved for \(I\).

Table of Integrals

A table of integrals is an incredibly handy reference tool that lists common integrals, much like a multiplication table lists the product of numbers. It enables students to find the antiderivatives of functions without going through the full integration process each time.

For elementary integrals such as \(\int \cos(u) du\), the table provides a direct answer: \(\sin(u) + C\). This direct approach saves time and helps verify results that might have been derived through more complex methodologies, such as integration by substitution or parts. Tables of integrals not only include simple cases but also integrals involving exponential, logarithmic, trigonometric, and power functions, among others.

It's important to note, however, that not all integrals are solved using tables, especially when they involve more intricate functions or require specific substitutions that alter the integrand significantly, as seen in our \(\cos(\ln x)\) example.