Chapter 5: Problem 363
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
Expert verified
\( \ln |\sin x - \cos x| + C \)
Step by step solution
01
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 \( du \).
02
Differentiation
Differentiate \( u \) with respect to \( x \): \( du = (\cos x + \sin x) dx \). This gives us a relationship between \( dx \) and \( du \).
03
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 \( \int \frac{1}{u} du \).
04
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.
05
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 \( x \).
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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:
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
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:
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 \( du = (\cos x + \sin x) dx \).
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:
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.