Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 8: Problem 25
Use the table of integrals at the back of the book to evaluate the integrals in Exercises \(1-26 .\) $$ \int \cos \frac{\theta}{3} \cos \frac{\theta}{4} d \theta $$
Short Answer
Expert verified
The integral is \( \frac{6}{7} \sin\left(\frac{7\theta}{12}\right) + 6 \sin\left(\frac{\theta}{12}\right) + C \).
Step by step solution
01
Understand the Problem
The problem requires us to evaluate the integral \( \int \cos \frac{\theta}{3} \cos \frac{\theta}{4} \, d\theta \) using a table of integrals. Our task is to find an equivalent integral formula from the table and apply it to our problem.
02
Use Trigonometric Identities
We need to express the integrand in a form that can be found in the integral table. One useful identity is the product-to-sum formulas.\[\cos A \cos B = \frac{1}{2}[\cos(A+B) + \cos(A-B)]\]Here, \(A = \frac{\theta}{3}\) and \(B = \frac{\theta}{4}\). Substituting these into the identity, we get:\[\cos \frac{\theta}{3} \cos \frac{\theta}{4} = \frac{1}{2}\left[\cos\left(\frac{\theta}{3}+\frac{\theta}{4}\right) + \cos\left(\frac{\theta}{3}-\frac{\theta}{4}\right)\right]\]
03
Simplify the Expression
Simplify the angles inside the cosine functions:\( \frac{\theta}{3} + \frac{\theta}{4} = \frac{4\theta}{12} + \frac{3\theta}{12} = \frac{7\theta}{12} \)\( \frac{\theta}{3} - \frac{\theta}{4} = \frac{4\theta}{12} - \frac{3\theta}{12} = \frac{\theta}{12} \)So our expression becomes:\[ \frac{1}{2}\left[\cos\left(\frac{7\theta}{12}\right) + \cos\left(\frac{\theta}{12}\right)\right] \]
04
Integrate Using Table of Integrals
Now we have:\[ \int \frac{1}{2}\left[\cos\left(\frac{7\theta}{12}\right) + \cos\left(\frac{\theta}{12}\right)\right] \,d\theta \]Using the integral formula \( \int \cos(k\theta) \,d\theta = \frac{1}{k} \sin(k\theta) + C \) from the table, we integrate:\[ \frac{1}{2}\left[\frac{12}{7} \sin\left(\frac{7\theta}{12}\right) + 12 \sin\left(\frac{\theta}{12}\right)\right] + C \]
05
Final Answer
Substitute back the expressions obtained:\[ \int \cos \frac{\theta}{3} \cos \frac{\theta}{4} \, d\theta = \frac{6}{7} \sin\left(\frac{7\theta}{12}\right) + 6 \sin\left(\frac{\theta}{12}\right) + C \]
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.
Trigonometric Identities
Trigonometric identities are crucial tools in simplifying and solving integrals involving trigonometric functions. These identities express relationships between trigonometric functions, making it easier to work with complicated expressions. For instance, the sum-to-product and product-to-sum identities help convert products of sine and cosine into sums that are easier to integrate or differentiate.
- The Pythagorean identities like \( \sin^2 \theta + \cos^2 \theta = 1 \), which help simplify expressions.
- Angle sum and difference formulas such as \( \sin(A \pm B) = \sin A \cos B \pm \cos A \sin B \) are beneficial for breaking down complex angles.
- The product-to-sum formula itself, \( \cos A \cos B = \frac{1}{2} [\cos(A+B) + \cos(A-B)] \), allows the conversion of products to sums, which is particularly useful in integration.
Product-to-Sum Formulas
The product-to-sum formulas are an essential set of trigonometric identities that convert products of sines and cosines into sums. These are particularly useful when dealing with integrals of products of trigonometric functions. The formula we used in the problem is:
\[ \cos A \cos B = \frac{1}{2} [\cos(A+B) + \cos(A-B)] \]This transformation is critical as integrals of sums are significantly easier to evaluate than products. In our given problem, using the product-to-sum formula transformed the integral into a manageable form, allowing us to use a standard integral table.
\[ \cos A \cos B = \frac{1}{2} [\cos(A+B) + \cos(A-B)] \]This transformation is critical as integrals of sums are significantly easier to evaluate than products. In our given problem, using the product-to-sum formula transformed the integral into a manageable form, allowing us to use a standard integral table.
- It reduces the complexity of expressions.
- Makes it easier to reference standard integral formulas.
- Facilitates algebraic manipulation of functions useful in calculus.
Table of Integrals
A table of integrals is a collection of formulas that provide the antiderivatives of numerous functions. It's a useful reference tool for quickly finding the integral of a function without performing the integration manually.
In our solution, the table of integrals gave us the formula for integrating \( \cos(k\theta) \), which is:
\[ \int \cos(k\theta) \,d\theta = \frac{1}{k} \sin(k\theta) + C \]This allowed us to apply the pre-calculated integral directly to the simplified expression, saving both time and effort.
In our solution, the table of integrals gave us the formula for integrating \( \cos(k\theta) \), which is:
\[ \int \cos(k\theta) \,d\theta = \frac{1}{k} \sin(k\theta) + C \]This allowed us to apply the pre-calculated integral directly to the simplified expression, saving both time and effort.
- These tables often provide antiderivatives for functions involving trigonometry, logarithms, and exponential forms.
- Using a table of integrals streamlines the integration process in complex problems.
- Ensures accuracy by relying on pre-verified solutions.
Definite Integral
Definite integrals are integrals with specified upper and lower limits shown as \( \int_{a}^{b} f(x) \, dx \). They not only compute the antiderivative but also evaluate the result at these limits to find the total 'net area' under a curve.
While the exercise did not directly deal with definite integrals, understanding them is essential:
While the exercise did not directly deal with definite integrals, understanding them is essential:
- They provide a method for calculating areas under the curve between two points.
- Used extensively in applications such as physics and engineering for finding work done, probability distributions, etc.
- The Fundamental Theorem of Calculus connects definite integrals with antiderivatives, making analytical evaluations possible.
Indefinite Integral
Indefinite integrals, as opposed to definite ones, don't have upper and lower bounds and are denoted as \( \int f(x) \, dx = F(x) + C \). Here, \( F(x) \) is the antiderivative of \( f(x) \), and \( C \) is the constant of integration.
The goal in an indefinite integral is to find a family of functions whose derivative gives the integrand:
The goal in an indefinite integral is to find a family of functions whose derivative gives the integrand:
- They represent the general form of antiderivatives without restricting to specific limits.
- Every solution differs by a constant \( C \), acknowledging any constant term in differentiation disappears.
- Used to determine the antiderivative as the first step before applying boundary conditions in real-world problems.
