Problem 59 - Calculus for Biology and Medicine

Question

Use the fact that $$ \cot x=\frac{\cos x}{\sin x} $$ to evaluate $$ \int \cot x d x $$

Step-by-Step Solution

Answer

The integral is \( \ln |\sin x| + C \).

Step 1: Rewrite cotangent using sine and cosine

The cotangent function can be rewritten as the ratio of cosine to sine. Thus, we start by replacing \( \cot x \) with \( \frac{\cos x}{\sin x} \) in the integral. This gives us the expression \( \int \frac{\cos x}{\sin x} \, dx \).

Step 2: Use substitution method

To solve the integral \( \int \frac{\cos x}{\sin x} \, dx \), we use substitution. Let \( u = \sin x \), which implies that \( du = \cos x \, dx \). This changes the integral to \( \int \frac{1}{u} \, du \).

Step 3: Integrate the transformed function

The integral \( \int \frac{1}{u} \, du \) is a standard logarithmic integral. The result of integrating \( \frac{1}{u} \) is \( \ln |u| + C \), where \( C \) is the constant of integration.

Step 4: Substitute back the original variable

Substitute \( u = \sin x \) back into the result from Step 3. This gives \( \ln |\sin x| + C \).

Key Concepts

Trigonometric Integrals

Trigonometric integrals involve the integration of functions that feature trigonometric expressions like sine, cosine, tangent, and their counterparts. These types of integrals are common, especially in calculus courses dealing with deeper integration techniques.

In the given exercise, we focus on integrating the cotangent function, which is one of these trigonometric expressions. Remember, the cotangent function is defined as the ratio of cosine to sine:

• \( \cot x = \frac{\cos x}{\sin x} \)

This representation is key because it allows us to simplify the integration process by reducing it to a more straightforward form.

Integrals involving trigonometric functions often require manipulation or transformation into forms that are easier or standard to integrate. Recognizing these forms is crucial as it simplifies the process of finding antiderivatives by making use of known integrals of basic functions.

Substitution Method

The substitution method is a powerful technique used to simplify the integration process, often turning a challenging integral into a much easier one. It involves replacing a specific part of the integrand with a single variable, commonly denoted by \( u \), that makes the integral easier to solve.

In this particular problem, the substitution method is used to integrate the function \( \int \frac{\cos x}{\sin x} \, dx \).

• First, we make a substitution by letting \( u = \sin x \).
• Then, find its differential \( du = \cos x \, dx \).
• This simplifies the integral into the form \( \int \frac{1}{u} \, du \).

The new integral is often easier to evaluate because it matches a basic formula that we know. The substitution method is especially useful for dealing with integrals containing complex algebraic, trigonometric, or exponential functions. It means converting these complexities into a single variable to ease integration.

Logarithmic Integration

Logarithmic integration refers to a specific type of integral involving the natural logarithm function, which arises when integrating functions like \( \frac{1}{x} \).

Once the substitution step has transformed our integral \( \int \frac{\cos x}{\sin x} \, dx \) into the simpler form \( \int \frac{1}{u} \, du \), it becomes a standard form of a logarithmic integral.

• The integral \( \int \frac{1}{u} \, du \) results in \( \ln |u| + C \).
• Here, \( C \) is the constant of integration, which must always be added to the result of an indefinite integral.

This result is a classic example of logarithmic integration, which frequently appears in calculus, especially when paired with trigonometric functions through substitution methods. Once the integral is evaluated, the final step involves substituting back the original variable we used: \( u = \sin x \). Thus, the solution for the integral is \( \ln |\sin x| + C \). This demonstrates not only the power of substitution but also how logarithmic functions help in solving integration problems with variable exponents or ratios.