Problem 363 - Calculus Volume 1

Question

In the following exercises, integrate using the indicated substitution. $$\int \frac{\sin x+\cos x}{\sin x-\cos x} d x ; u=\sin x-\cos x$$

Step-by-Step Solution

Answer

\( \ln |\sin x - \cos x| + C \)

Step 1: Substitution

We are asked to use the substitution method. Here, we define the substitution: let \( u = \sin x - \cos x \). Differentiate both sides to find \( du \).

Step 2: Differentiation

Differentiate \( u \) with respect to \( x \): \( du = (\cos x + \sin x) dx \). This gives us a relationship between \( dx \) and \( du \).

Step 3: Rewrite the Integral

Substitute \( u \) and \( du \) in the integral. The integral becomes: \( \int \frac{\sin x + \cos x}{u} d x \). Substitute \( du = (\cos x + \sin x) dx \), so \( dx = \frac{du}{\cos x + \sin x} \). The integral becomes \( \int \frac{1}{u} du \).

Step 4: Integrate

The integral \( \int \frac{1}{u} du \) is a standard natural logarithm integral. Thus, the integral of \( 1/u \) is \( \ln |u| + C \), where \( C \) is the constant of integration.

Step 5: Back Substitute

Now substitute back \( u = \sin x - \cos x \), which gives us: \( \ln |\sin x - \cos x| + C \). This is the solution in terms of \( x \).

Key Concepts

Trigonometric Integration

Trigonometric integration is a technique used to solve integrals involving trigonometric functions like \( \sin x \), \( \cos x \), \( \tan x \), and others.
It often involves clever substitutions or manipulations based on trigonometric identities. In this exercise, the integral involves both \( \sin x \) and \( \cos x \), which are fundamental trigonometric functions.
A common approach in trigonometric integration is to simplify the expression using trigonometric identities, such as \( \sin^2 x + \cos^2 x = 1 \). However, in this exercise, the solution uses a substitution method to enable integration.
This shows that sometimes simply exchanging or rewriting in terms of another function can pave the way to a straightforward solution.
The main objective in any trigonometric integration problem is to transform the given expression into a form that is easier to integrate directly. This usually involves:

• Identifying parts of the integral that fit trigonometric identities
• Using substitutions to simplify the integral
• Rewriting the expression to match known integral forms

Overall, while trigonometric integration can be seen as complex, most solutions involve strategic manipulation of functions to reach a simpler form.

Differentiation in Calculus

Differentiation in calculus is a key concept used to find the rate at which a function is changing at any point. It forms the backbone of techniques used in integration by substitution.
In the provided exercise, differentiation was performed on the substitution \( u = \sin x - \cos x \) to find \( du \).
Differentiating this expression gives us \( du = (\cos x + \sin x) dx \). This step is crucial because it transforms the original variable \( x \) into the new variable \( u \), allowing us to rewrite the integral in terms of \( u \).
This process of creating \( du \) is what truly simplifies the integral.
The following happens during differentiation:

• Identify the function you need to differentiate.
• The derivative of \( \sin x \) is \( \cos x \), and the derivative of \( -\cos x \) is \(-\sin x \).
• Combine these results to arrive at \( du = (\cos x + \sin x) dx \).

Understanding differentiation is vital because it allows us to convert complicated integrals into simpler forms, facilitating their solution.

Natural Logarithm Integration

Natural logarithm integration involves finding the integral of functions of the form \( \frac{1}{u} \), where \( u \) is a variable or function. The formula \( \int \frac{1}{u} du = \ln |u| + C \) is a foundational tool when solving these integrals.
In this exercise, after substitution, the integral simplifies to \( \int \frac{1}{u} du \).
When solving for the integral \( \int \frac{1}{u} du \), we recognize immediately that it results in \( \ln |u| + C \), where \( C \) represents the constant of integration. This is because the natural logarithm is the antiderivative of \( \frac{1}{u} \).
Remember, the absolute value in \( \ln |u| \) is necessary since the logarithm of a negative number is undefined in the real numbers.
Points to consider while applying natural logarithm integration:

• The form \( \frac{1}{u} \) results directly in natural logarithm form upon integration.
• Always account for absolute values to maintain the domain of the logarithm function.
• Don't forget the constant \( C \), which represents any constant value that could have existed before differentiation.

This integration method is efficient and indispensable for solving many real-world problems and mathematical models that involve logarithmic relations.