Problem 14 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 1-28, perform the indicated integrations. \(\int \cos y \cos 4 y d y\)

Step-by-Step Solution

Answer

\(\frac{1}{10} \sin 5y + \frac{1}{6} \sin 3y + C\)

Step 1: Use Product-to-Sum Identities

The integral involves a product of cosine functions. To integrate this, we use the product-to-sum identities.Recall the identity: \[\cos A \cos B = \frac{1}{2}(\cos(A + B) + \cos(A - B))\]For our problem, let \(A = y\) and \(B = 4y\). Substitute these into the identity:\[\cos y \cos 4y = \frac{1}{2}\left(\cos(y + 4y) + \cos(y - 4y)\right) = \frac{1}{2}\left(\cos 5y + \cos(-3y)\right)\]Since \(\cos(-3y) = \cos 3y\), the expression simplifies to:\[\cos y \cos 4y = \frac{1}{2}\left(\cos 5y + \cos 3y\right)\]

Step 2: Set Up the Integral

Substitute the expression from Step 1 into the integral:\[\int \cos y \cos 4y \, dy = \int \frac{1}{2}\left(\cos 5y + \cos 3y\right) \, dy\]This gives us two separate integrals:\[= \frac{1}{2} \int \cos 5y \, dy + \frac{1}{2} \int \cos 3y \, dy\]

Step 3: Integrate Each Term Separately

Let's integrate each term separately.For the first integral:\[\int \cos 5y \, dy = \frac{1}{5} \sin 5y + C_1\]For the second integral:\[\int \cos 3y \, dy = \frac{1}{3} \sin 3y + C_2\]

Step 4: Combine Results

Combine the results of the separate integrals, including the \(\frac{1}{2}\) factor:\[\frac{1}{2} \left( \frac{1}{5} \sin 5y + \frac{1}{3} \sin 3y \right)\]Distribute the \(\frac{1}{2}\):\[= \frac{1}{10} \sin 5y + \frac{1}{6} \sin 3y + C\] where \(C\) is a constant representing the sum of integration constants \(C_1\) and \(C_2\).

Step 5: Final Answer

The integral\[\int \cos y \cos 4y \, dy\] evaluates to:\[\frac{1}{10} \sin 5y + \frac{1}{6} \sin 3y + C\]

Key Concepts

product-to-sum identities

Understanding product-to-sum identities is crucial when dealing with integrals of trigonometric functions. These identities allow us to transform products of trigonometric functions into sums, which are often easier to integrate. The product-to-sum identity for cosine functions is particularly useful:

• \( \cos A \cos B = \frac{1}{2}(\cos(A + B) + \cos(A - B)) \)

This formula helps in simplifying expressions involving multiplications of functions like \( \cos y \cos 4y \). Substituting values \( A = y \) and \( B = 4y \), the identity converts the product into a sum:

• \( \cos y \cos 4y = \frac{1}{2}(\cos 5y + \cos 3y) \)

Simplifying the problem this way is the first step in making the integral more manageable. These identities are powerful tools, as they turn complex integrals into a combination of simpler ones.

trigonometric integrals

Trigonometric integrals involve expressions containing trigonometric functions like sine, cosine, tangent, etc. These types of integrals can be tricky due to their oscillatory nature. However, the transformation using product-to-sum identities simplifies the integration significantly, breaking down a complex integral into parts that are straightforward.After transforming the product \( \cos y \cos 4y \) using product-to-sum identities, we arrive at:

• \( \int \frac{1}{2}(\cos 5y + \cos 3y) \, dy \)

The integral now splits into separate integrals for each cosine term, namely:

• \( \int \cos 5y \, dy \) and \( \int \cos 3y \, dy \)

Each of these can be solved using basic integration techniques for trigonometric functions, giving us:

• \( \int \cos 5y \, dy = \frac{1}{5} \sin 5y + C_1 \)
• \( \int \cos 3y \, dy = \frac{1}{3} \sin 3y + C_2 \)

Breaking down the problem like this makes the process systematic and reduces errors.

calculus problem solving

Problem solving in calculus involves a strategic approach to tackling various kinds of problems effectively. Integrating trigonometric functions such as \( \int \cos y \cos 4y \, dy \) highlights several key tactics:

• Identify familiar patterns or identities that can simplify the problem, like product-to-sum identities.
• Break down the problem into smaller, more manageable parts. This involves separating the integral based on the simplified expression.
• Use known integration techniques. For instance, integrating terms like \( \cos 5y \, dy \) involves applying standard results for integrals of cosine to obtain \( \frac{1}{5} \sin 5y \).

Once each part is solved, combine the results correctly, ensuring all constants from integration are included. The final solution for trigonometric integrals often includes a constant \( C \), representing the sum of constants from each part of integration.Overall, proficiency in using calculus techniques requires practice and familiarity with different expressions and identities, which enhance problem-solving skills.