Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 16: Problem 27
Evaluate each integral. $$\int \sin (7 x-3) d x$$
Short Answer
Expert verified
The integral is \(-\frac{1}{7} \cos(7x-3) + C\).
Step by step solution
01
Identify the Integral Type
The integral we need to solve is \( \int \sin(7x - 3) \, dx \). This is a basic sine integral, which can be solved using a substitution method or recognizing its standard form.
02
Determine the Substitution
Let \( u = 7x - 3 \). Then the derivative of \( u \, \) with respect to \( x \, \) is \( du/dx = 7 \). Therefore, \( dx = du/7 \).
03
Substitute and Rewrite the Integral
Substitute \( u = 7x - 3 \) into the integral, which gives us \( \int \sin(u) \, (du/7) \). This can be simplified to \( \frac{1}{7} \int \sin(u) \, du \).
04
Integrate using Standard Form
The standard integral of \( \sin(u) \) is \( -\cos(u) \). So, \( \int \sin(u) \, du = -\cos(u) \). Apply this to get \( \frac{1}{7} (-\cos(u)) = -\frac{1}{7} \cos(u) \).
05
Substitute Back to Original Variable
Replace \( u \) with \( 7x - 3 \) to get the solution in terms of \( x \). Therefore, the integral is \( -\frac{1}{7} \cos(7x - 3) + C \), where \( C \) is the constant of integration.
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 91影视!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Substitution Method
To solve integrals like \( \int \sin(7x - 3) \, dx \), we often use the substitution method. This technique simplifies the integral by changing variables to one that is easier to manage. Here鈥檚 how it works in this context:
- First, identify a part of the integral to substitute. For \( \int \sin(7x - 3) \, dx \), we choose \( u = 7x - 3 \).
- Next, find the derivative \( \frac{du}{dx} \), which is 7 in this case. This implies \( du = 7\, dx \), or rearranging gives us \( dx = \frac{du}{7} \).
- Substituting, the integral \( \int \sin(u) \, \frac{du}{7} \) simplifies the process because it changes the focus from handling \( 7x - 3 \) as \( u \).
- Always remember to adjust for \( dx \), converting it into \( du \) as part of the substitution.
Sine Function
The sine function is fundamental in trigonometry and plays a central role in calculus as well. When you're facing an integral like \( \int \sin(7x - 3) \, dx \), understanding its properties streamlines the integration process.
- The sine function, denoted as \( \sin(x) \), oscillates between -1 and 1, creating a smooth wave pattern.
- Because of its wavy nature, the integral of the sine function, \( \int \sin(x) \, dx \), is \( -\cos(x) \). This relies on the relationship that differentiation reverses the integration; differentiating \( -\cos(x) \) returns \( \sin(x) \).
- In our problem, employing substitution made sure \( \int \sin(7x-3) \, dx \) becomes \( \int \sin(u) \, du \). Thus, it directly uses the standard form \( -\cos(u) \).
Standard Integral Form
One of the conveniences in calculus is recognizing standard integral forms. For \( \int \sin(u) \, du \), there's a direct formula that can be applied:
- The standard integration form for sine is \( \int \sin(u) \, du = -\cos(u) + C \), where \( C \) denotes the constant of integration.
- This form simplifies the integration of any sine function, transforming problems into easier calculations once the variable \( u \) is consistent.
- After applying substitution in our example, the integral becomes \( \frac{1}{7} \int \sin(u) \, du \). Use the standard form to solve it, leading to \( -\frac{1}{7} \cos(u) + C \).
- Finally, revert \( u \) to the original variable, balancing the solution as \( -\frac{1}{7} \cos(7x - 3) + C \). This step is essential to finalizing the integration in terms of the original equation.
