Problem 13
Question
Perform the indicated integrations. $$ \int \sin 4 y \cos 5 y d y $$
Step-by-Step Solution
Verified Answer
The integral is \( -\frac{1}{18} \cos 9y + \frac{1}{2} \cos y + C \).
1Step 1: Identify the Strategy
Since the integrand involves the product of \( \sin 4y \) and \( \cos 5y \), we can apply the product-to-sum identities. The identity \( \sin A \cos B = \frac{1}{2} [\sin(A + B) + \sin(A - B)] \) is particularly useful in this case.
2Step 2: Apply the Product-to-Sum Identity
Using the identity from Step 1, we have \( \sin 4y \cos 5y = \frac{1}{2} [\sin(4y + 5y) + \sin(4y - 5y)] = \frac{1}{2} [\sin 9y + \sin(-y)] \). Since \( \sin(-y) = -\sin y \), the expression simplifies to \( \frac{1}{2} [\sin 9y - \sin y] \).
3Step 3: Set Up the New Integral
Substitute the expression from Step 2 into the integral: \( \int \sin 4y \cos 5y \, dy = \int \frac{1}{2} [\sin 9y - \sin y] \, dy \). This can be split into two separate integrals: \( \frac{1}{2} \int \sin 9y \, dy - \frac{1}{2} \int \sin y \, dy \).
4Step 4: Integrate Each Term
First, integrate \( \sin 9y \): The integral of \( \sin(kx) \) is \(-\frac{1}{k} \cos(kx) \). Here \( k = 9 \), so \( \int \sin 9y \, dy = -\frac{1}{9} \cos 9y + C_1 \). For \( \sin y \), the integral is \(-\cos y + C_2 \).
5Step 5: Combine the Results
Substitute the integrals from Step 4 back into the expression: \( \frac{1}{2} \left( -\frac{1}{9} \cos 9y + C_1 \right) - \frac{1}{2} (-\cos y + C_2) \). Simplify this to obtain the final result: \( -\frac{1}{18} \cos 9y + \frac{1}{2} \cos y + C \), where \( C \) is a constant that combines \( C_1 \) and \( C_2 \).
Key Concepts
Trigonometric IdentitiesProduct-to-Sum FormulasAntiderivativesIntegration Techniques
Trigonometric Identities
Trigonometric identities are fundamental in calculus, especially when dealing with integration problems involving trigonometric functions. These are equations that are true for all values of the variables involved. In our example, the integral \( \int \sin 4y \cos 5y \, dy \) involves products of sine and cosine. A basic understanding of identities, such as \( \sin^2 x + \cos^2 x = 1 \), allows for the manipulation and simplification of functions to make integration possible.
\( \sin(A \pm B), \cos(A \pm B) \) can often be rewritten in simpler forms using identities. This broadens the scope of functions we can integrate by traditional means. Knowing and applying these identities quickly transforms complex integrals into manageable ones.
\( \sin(A \pm B), \cos(A \pm B) \) can often be rewritten in simpler forms using identities. This broadens the scope of functions we can integrate by traditional means. Knowing and applying these identities quickly transforms complex integrals into manageable ones.
Product-to-Sum Formulas
Product-to-sum formulas are invaluable tools in integrating products of trigonometric functions. These formulas, such as \( \sin A \cos B = \frac{1}{2}[\sin(A + B) + \sin(A - B)] \), allow us to express products of sines and cosines as sums, which are simpler to integrate.
In the given exercise, this technique converts the original integrand \( \sin 4y \cos 5y \) into \( \frac{1}{2} [\sin 9y - \sin y] \). This transformation simplifies the integration process by changing a product of functions into a linear combination of simpler functions.
Becoming familiar with these formulas is crucial for efficiently tackling integrations that involve multiple trigonometric terms. They reduce the complexity by transforming the integrals into basic sine and cosine functions.
In the given exercise, this technique converts the original integrand \( \sin 4y \cos 5y \) into \( \frac{1}{2} [\sin 9y - \sin y] \). This transformation simplifies the integration process by changing a product of functions into a linear combination of simpler functions.
Becoming familiar with these formulas is crucial for efficiently tackling integrations that involve multiple trigonometric terms. They reduce the complexity by transforming the integrals into basic sine and cosine functions.
Antiderivatives
Antiderivatives form the backbone of integration; they provide a function whose derivative matches the given integrand. For trigonometric functions, understanding their antiderivatives is key to solving the integral.For example, the antiderivative of \( \sin(kx) \) is \(-\frac{1}{k}\cos(kx) + C\). In our solution, knowing the antiderivatives allows us to integrate each term separately:
- \( \int \sin 9y \, dy = -\frac{1}{9} \cos 9y + C_1 \)
- \( \int \sin y \, dy = -\cos y + C_2 \)
Integration Techniques
In calculus, integration techniques are essential for solving integrals effectively. These techniques are varied, catering to different forms of functions. Familiarity with them aids in simplifying complex integrals. By incorporating techniques such as substitution, by-parts, and trigonometric identities, we tackle the integral efficiently. In our case, using the product-to-sum formula is a technique that simplifies the expression \( \sin 4y \cos 5y \) initially.After applying this transformation, breaking the integral into manageable parts and solving each allows us to determine the integral quickly:
- Split the problem into simple known integrals
- Integrate each part independently
- Combine results and simplify
Other exercises in this chapter
Problem 13
Solve each differential equation. $$ x y^{\prime}+(1+x) y=e^{-x} ; y=0 \text { when } x=1 \text { . } $$
View solution Problem 13
Use the method of partial fraction decomposition to perform the required integration. \(\int \frac{2 x^{2}+x-4}{x^{3}-x^{2}-2 x} d x\)
View solution Problem 13
Perform the indicated integrations. $$ \int \frac{\sin \sqrt{t}}{\sqrt{t}} d t $$
View solution Problem 13
$$ \text { use integration by parts to evaluate each integral. } $$ $$ \int \arctan x d x $$
View solution