Problem 90 - Calculus Early Transcendentals

Question

Looking ahead: Integrals of \(\tan x\) and cot \(x\) Use a change of variables to verify each integral. a. \(\int \tan x d x=-\ln |\cos x|+C=\ln |\sec x|+C\) b. \(\int \cot x d x=\ln |\sin x|+C\)

Step-by-Step Solution

Answer

Question: Prove the following integral formulas using a change of variables: a) \(\int \tan x dx = -\ln |\cos x|+C=\ln |\sec x|+C\), and b) \(\int \cot x dx= \ln |\sin x|+C\). Answer: a) We used the substitution \(u = \cos x\) and obtained \(-\ln|\cos x|+C = \ln|\sec x|+C\), proving that \(\int \tan x dx = -\ln |\cos x|+C=\ln |\sec x|+C\). b) We used the substitution \(u = \sin x\) and found the integral to be \(\ln|\sin x|+C\), proving that \(\int \cot x dx= \ln |\sin x|+C\).

Step 1: Setting Up the Substitution

Let \(u = \cos x\). Then, we have \(du = -\sin x dx\).

Step 2: Substituting

Now substitute \(u\) in the integral, we get \(\int \frac{\sin x}{\cos x} dx = \int \frac{1}{u}(-du) = - \int \frac{1}{u} du\).

Step 3: Evaluating the Integral

We know that the integral of \(\frac{1}{u}\) is \(\ln|u|+C\). So, \(- \int \frac{1}{u} du = -\ln|u|+C = -\ln|\cos x|+C\).

Step 4: Converting to secant

Recall that \(\sec x = \frac{1}{\cos x}\). Therefore, \(-\ln|\cos x|+C=\ln|\sec x|+C\). So, we have proved that \(\int \tan x dx = -\ln|\cos x|+C=\ln|\sec x|+C\). #b. \(\int \cot x dx= \ln |\sin x|+C\)#

Step 1: Setting Up the Substitution

Let \(u = \sin x\). Then, we have \(du = \cos x dx\).

Step 2: Substituting

Now substitute \(u\) in the integral, we get \(\int \frac{\cos x}{\sin x} dx = \int \frac{1}{u} du\).

Step 3: Evaluating the Integral

We know that the integral of \(\frac{1}{u}\) is \(\ln|u|+C\). So, \(\int \frac{1}{u} du = \ln|u|+C = \ln|\sin x|+C\).

Step 4: Final Result

Thus, we have verified that \(\int \cot x dx= \ln |\sin x|+C\).

Key Concepts

Trigonometric Integrals

Trigonometric integrals involve the integration of products or powers of trigonometric functions. These integrals are fundamental in calculus, especially when dealing with periodic functions. In the given exercise, we work with the integrals of tangent \(\tan x\) and cotangent \(\cot x\). These function integrals are common in many calculus problems.

• The integral of \(\tan x\) is \(\int \tan x \, dx = -\ln |\cos x| + C = \ln |\sec x| + C\).
• The integral of \(\cot x\) is \(\int \cot x \, dx = \ln |\sin x| + C\).

These results are crucial as they often arise in trigonometric substitution problems. Understanding how to integrate these functions strengthens foundational calculus skills.
Trigonometric integrals often use identities and substitutions to simplify expressions before integrating. By being familiar with these concepts, one can solve more complex integration problems effectively.

Substitution Method

The substitution method is a fundamental technique in integral calculus. It's used to simplify integrals by making a change of variables. This method is especially useful for trigonometric integrals, as seen in the exercise. Here's a step-by-step overview of how it's applied:

Substitution for \(\tan x\):

• Step 1: Choose a substitution. Let \(u = \cos x\).
• Step 2: Differentiate. Find \(du = -\sin x \, dx\).
• Step 3: Rewrite the integral. Substitute \(u\) and \(du\) into the integral, resulting in \(\int \frac{1}{u}(-du)\).
• Step 4: Integrate in terms of \(u\). The integral becomes \(-\ln |u| + C\), which translates back to the original variable as \(-\ln |\cos x| + C\).

Substitution for \(\cot x\):

• Step 1: Choose a substitution. Let \(u = \sin x\).
• Step 2: Differentiate. Find \(du = \cos x \, dx\).
• Step 3: Rewrite the integral. Substitute \(u\) and the integral becomes \(\int \frac{1}{u} \, du\).
• Step 4: Integrate in terms of \(u\). The integral is \(\ln |u| + C\), returning to \(\ln |\sin x| + C\) in terms of \(x\).

The substitution method relies on finding a suitable substitution that simplifies the integral, writing the integral in terms of a new variable, and then integrating. This method not only simplifies the integration process but also provides deeper insight into the structure of the function being integrated.

Indefinite Integrals

Indefinite integrals represent a family of functions and are expressed with a constant of integration, noted by \(C\). Understanding indefinite integrals is key to mastering integral calculus as they provide general solutions to differential equations.

• An indefinite integral has the form \( \int f(x) \, dx = F(x) + C,\) where \(F(x)\) is the antiderivative of \(f(x)\).

In the exercise, after applying substitutions and solving the integral, we obtain results like \(\int \tan x \, dx = \ln |\sec x| + C\) and \(\int \cot x \, dx = \ln |\sin x| + C\).
These results showcase how indefinite integrals manifest as broader expressions, inclusive of any number of possible constant shifts represented by \(C\).

• This constant accounts for the indefinite nature, allowing any vertical shift of the antiderivative, which is essential in finding particular solutions to problems.
• A key aspect is interpreting the absolute value within the logarithms, which ensures the expressions remain valid over all appropriate domains of \(x\).

Mastery of indefinite integrals opens doors to solving various kinds of calculus problems, from basic equations to complex differential scenarios.