Problem 16 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 1-28, perform the indicated integrations. \(\int \sin 3 t \sin t d t\)

Step-by-Step Solution

Answer

\(\frac{1}{4} \sin 2t - \frac{1}{8} \sin 4t + C\)

Step 1: Identify the Product to Use a Formula

The integral \( \int \sin 3t \sin t \, dt \) involves the product of two sine functions. We will use the product-to-sum identities for sine: \( \sin A \sin B = \frac{1}{2} [\cos(A-B) - \cos(A+B)] \).

Step 2: Apply the Product-to-Sum Identity

Using the identity, substitute \( A = 3t \) and \( B = t \) into the formula: \( \sin 3t \sin t = \frac{1}{2}[\cos(3t-t) - \cos(3t+t)] = \frac{1}{2}[\cos(2t) - \cos(4t)] \).

Step 3: Rewrite the Integral

Substitute the identity result back into the integral: \( \int \sin 3t \sin t \, dt = \int \frac{1}{2}[\cos 2t - \cos 4t] \, dt \). This simplifies to \( \frac{1}{2} \left( \int \cos 2t \, dt - \int \cos 4t \, dt \right) \).

Step 4: Integrate Each Term Separately

Start by integrating \( \cos 2t \): \( \int \cos 2t \, dt = \frac{1}{2} \sin 2t + C \). Next, integrate \( \cos 4t \): \( \int \cos 4t \, dt = \frac{1}{4} \sin 4t + C \).

Step 5: Substitute Integrals Back into Expression

Substitute the results of the integrals back into the expression: \( \frac{1}{2} \left( \frac{1}{2} \sin 2t - \frac{1}{4} \sin 4t \right) \).

Step 6: Simplify the Expression

Multiply through by \( \frac{1}{2} \): \( \frac{1}{4} \sin 2t - \frac{1}{8} \sin 4t + C \). This is the final integration result.

Key Concepts

Trigonometric Identities

Trigonometric identities are crucial in simplifying expressions involving trigonometric functions. In this exercise, we deal with identities that help transform products of trigonometric functions into sums or differences, making integration simpler. Remember:

• Simplify complex expressions: They often allow us to express complex trigonometric functions in a more practical form.
• Common identities: Include Pythagorean, angle sum, and double angle identities, which all have unique uses depending on the problem.
• Transformation: In this example, using the identity for the product of two sines helps convert the function into a form that's much easier to integrate directly.

Understanding these identities is the key to solving more challenging calculus problems with ease.

Product-to-Sum Formula

The product-to-sum formula is an incredibly useful tool in calculus, especially when dealing with the integration of products of trigonometric functions. This formula allows us to rewrite the product of two sines or cosines as a sum or difference of cosines. Here’s the formula used:

• For two sine functions: \( \sin A \sin B = \frac{1}{2} [\cos(A-B) - \cos(A+B)] \)

In this exercise:

• Substitute correctly: With \( A = 3t \) and \( B = t \), we apply the formula to transform \( \sin 3t \sin t \) into \( \frac{1}{2} [\cos 2t - \cos 4t] \).
• Easier integration: This transformation simplifies the process of integration drastically.

Mastering the application of these formulas not only simplifies the process but also prevents errors in transitioning between steps.

Definite Integrals

Definite integrals are not directly part of this exercise, but understanding them enriches the knowledge required to apply integration techniques. Unlike indefinite integrals, definite integrals compute the area under the curve between two specific points on the x-axis. Key features include:

• Boundaries: Defined by upper and lower limits, which specify the interval of integration.
• Area calculation: Helps in finding exact values like area under a curve or between curves.
• Applications: Used in physics for finding displacement, in economics for total cost, and other fields where accumulation over an interval is required.

While this exercise focuses on indefinite integrals, the principles of integration remain the same, making it a stepping stone to solving definite integrals.

Calculus

Calculus, the mathematics of change, encompasses techniques for integration and differentiation—two of its primary operations. This exercise specifically focuses on integration, which is used for:

• Antiderivatives: Finding a function whose derivative is the given function.
• Area under curves: Essential in fields like physics and engineering, integration helps calculate the total quantity accreted over time or space.
• Key concept: Involves breaking down complex functions into simpler parts that are more manageable to integrate, evident in the use of trigonometric identities and transformation formulas.

Calculus is widely applicable and mastering it allows for solving a plethora of real-world problems, from calculating growth rates to optimization in various disciplines.