Question

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

Short answer

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

Step-by-step solution

Identify the Substitution

We are asked to perform a substitution with \( u = \sin x - \cos x \). This means we will change variables from \( x \) to \( u \).

Differentiate for du

To find \( du \), we differentiate \( u \) with respect to \( x \). Therefore, \( \frac{du}{dx} = \cos x + \sin x \). Rearranging gives \( du = (\cos x + \sin x) \, dx \).

Express dx in terms of du

From the equation \( du = (\cos x + \sin x) \, dx \), solve for \( dx \) to substitute it back into the integral. Thus, \( dx = \frac{du}{\cos x + \sin x} \).

Substitute into the Integral

Replace \( \sin x - \cos x \) with \( u \) in the integral, and substitute \( dx = \frac{du}{\cos x + \sin x} \). The integral becomes: \[ \int \frac{\sin x + \cos x}{u} \cdot \frac{du}{\cos x + \sin x}, \] which simplifies to \[ \int \frac{1}{u} \, du. \]

Integrate

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

Back-Substitute for u

Finally, replace \( u \) with the original substitution \( \sin x - \cos x \). Therefore, the integral is \( \ln |\sin x - \cos x| + C \).

Key concepts

Differentiation

Differentiation is a fundamental concept in calculus that helps us understand how a function changes at any given point. It involves finding the derivative of a function, which is a measure of the rate at which the function's value changes.

In the context of this exercise, differentiation is used to find the derivative of a substitution variable. When substituting variables in integrals, like our example where we use the substitution \( u = \sin x - \cos x \), we need to calculate \( \frac{du}{dx} \). To do this, we differentiate \( u \) with respect to \( x \), yielding \( \frac{du}{dx} = \cos x + \sin x \).

The result allows us to express \( du \) as \( du = (\cos x + \sin x) \, dx \), which is crucial for substituting \( dx \) in the integral. Differentiation is key because it provides the relationship between the new variable \( u \) and the original variable \( x \), allowing us to transform and simplify the integral.

Definite integrals

A definite integral is a powerful tool in calculus used to calculate the area under a curve within specified bounds. Unlike an indefinite integral, which represents a family of functions, a definite integral provides a single numeric value.

In our exercise, although we are working with an indefinite integral, understanding definite integrals is crucial, as it is the next step in solving many problems. The process we followed with substitution makes integrals easier to handle, transforming them into simpler forms that we can evaluate more straightforwardly.

• A definite integral has limits (bounds), usually written as \( \int_{a}^{b} f(x) \, dx \), representing the area from \( x = a \) to \( x = b \).
• This area corresponds to the cumulative sum of infinitesimally small rectangles under the curve \( f(x) \).
• Evaluating these integrals provides insights into real-world problems, such as computing distance, area, and even probabilities.

Understanding the concept of definite integrals prepares one for applying these techniques to real scenarios and more complex integrals.

Inverse trigonometric functions

Inverse trigonometric functions are the inverses of the basic trigonometric functions such as sine, cosine, and tangent. They allow us to find angles when given function values.

Though not directly used in our given exercise, understanding inverse trigonometric functions can be crucial when working with integrals and derivatives that involve trigonometric expressions. They often appear when reversing trigonometric functions back to their initial angles.

For example, if we have \( \sin^{-1}(x) \), it represents the angle whose sine is \( x \). Knowing these functions helps in solving integrals that arise from transformations involving trigonometric identities or substitutions, particularly when results need to be expressed in terms of angles rather than numerical values.

• Such functions include \( \sin^{-1}(x) \), \( \cos^{-1}(x) \), and \( \tan^{-1}(x) \).
• They can simplify the process of integrating certain complex expressions related to trigonometric identities.
• Understanding these can broaden the scope of problem-solving techniques available to students.

By becoming familiar with inverse trigonometric functions, one gains a complete toolkit for handling a variety of trigonometric-related mathematical problems efficiently.