Problem 60 - Thomas Calculus

Question

Evaluate each integral in Exercises \(57-62\) by multiplying by a form of 1 and using a substitution (if necessary) to reduce it to standard form. $$ \int \frac{1}{\csc \theta+\cot \theta} d \theta $$

Step-by-Step Solution

Answer

The integral evaluates to \( -\ln |1 + \cos \theta| + C \).

Step 1: Simplify Trigonometric Expression

We want to simplify the integrand \( \frac{1}{\csc \theta + \cot \theta} \). Recall the identity \( \csc \theta = \frac{1}{\sin \theta} \) and \( \cot \theta = \frac{\cos \theta}{\sin \theta} \). The expression becomes \( \csc \theta + \cot \theta = \frac{1}{\sin \theta} + \frac{\cos \theta}{\sin \theta} = \frac{1 + \cos \theta}{\sin \theta} \). Thus, the integrand is \( \frac{\sin \theta}{1 + \cos \theta} \).

Step 2: Use Substitution to Simplify the Integral

Set \( u = 1 + \cos \theta \). Then \( \frac{du}{d\theta} = -\sin \theta \), so \( du = -\sin \theta \, d\theta \) or \( -du = \sin \theta \, d\theta \). Substituting into the integral gives \( \int \frac{\sin \theta}{1 + \cos \theta} d\theta = \int \frac{\sin \theta}{u} \cdot d\theta = \int \frac{-du}{u} \). This simplifies to \( -\int \frac{du}{u} \).

Step 3: Evaluate the Integral

The integral \( -\int \frac{du}{u} \) is a standard logarithmic integral. It evaluates to \( -\ln |u| + C \), where \( C \) is the constant of integration.

Step 4: Substitute Back the Original Variable

Substitute back \( u = 1 + \cos \theta \) into the result. We obtain \( -\ln |1 + \cos \theta| + C \).

Key Concepts

Trigonometric Identities

Trigonometric identities are essential tools in calculus for simplifying expressions and solving integrals. In this context, recognizing the identities for cosecant and cotangent is crucial.

• The identity for cosecant is: \( \csc \theta = \frac{1}{\sin \theta} \).
• Cotangent is expressed as: \( \cot \theta = \frac{\cos \theta}{\sin \theta} \).

We use these identities to simplify complex trigonometric expressions. For instance, in the expression \( \csc \theta + \cot \theta \), applying these identities allows us to combine and reduce the computation to a more manageable form: \( \frac{1 + \cos \theta}{\sin \theta} \). Understanding and applying trigonometric identities makes integrations and transformative algebra more intuitive and precise.

Substitution Method

Substitution is a powerful technique in calculus that simplifies complex integrals. It's like a change of variables that transforms a difficult problem into one we can handle more easily. In our exercise, we utilize substitution to manage the integral of \( \frac{\sin \theta}{1 + \cos \theta} \).To use substitution, we set \( u = 1 + \cos \theta \). This step requires computing the differential \( du \), which helps in reshaping the integral's differentials:

• First, find the derivative of \( u \) with respect to \( \theta \): \( \frac{du}{d\theta} = -\sin \theta \).
• Thus, \( du = -\sin \theta \, d\theta \) or equivalently, \( -du = \sin \theta \, d\theta \).

By making this substitution, our integral simplifies to \( \int \frac{-du}{u} \). This process reduces a potentially convoluted integral into a simpler form, making it easier to integrate.

Logarithmic Integration

Logarithmic integration is a method used when integrating functions of the form \( \int \frac{1}{u} \, du \), which results in a natural logarithm. Applying this to our substituted integral, \( \int \frac{-du}{u} \), is straightforward.The integral of \( \frac{1}{u} \) is \( \ln |u| \), so when there's a negative sign, it becomes:

• \( -\int \frac{du}{u} = -\ln |u| + C \),

where \( C \) is the constant of integration. Finally, we revert to the original variable by substituting back \( u = 1 + \cos \theta \) into our expression, yielding the solution \( -\ln |1 + \cos \theta| + C \). This logarithmic result neatly solves our integral in terms of the initial trigonometric expression.