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Chapter 8: Problem 52

Evaluate the integrals in Exercises \(51-56\) $$ \int \sin 2 x \cos 3 x d x $$

Short Answer

Expert verified
The integral evaluates to \( -\frac{1}{10}\cos(5x) + \frac{1}{2}\cos(x) + C \).

Step by step solution

01

Use Product-to-Sum Formulas

The integral \( \int \sin 2x \cos 3x \ dx \) can be simplified using the product-to-sum formulas. One of these formulas is \( \sin A \cos B = \frac{1}{2} [\sin(A+B) + \sin(A-B)] \). Using this, we rewrite the integrand: \( \sin 2x \cos 3x = \frac{1}{2} [\sin(5x) + \sin(-x)] \). Since \( \sin(-x) = -\sin(x) \), it simplifies further to \( \frac{1}{2} [\sin(5x) - \sin(x)] \).
02

Split the Integral

Given our expression \( \frac{1}{2} [\sin(5x) - \sin(x)] \), we split the integral into two separate integrals: \( \int \frac{1}{2} \sin(5x) \ dx - \int \frac{1}{2} \sin(x) \ dx \).
03

Integrate Each Term Separately

Start by integrating \( \int \frac{1}{2} \sin(5x) \ dx \). The integral of \( \sin(kx) \) is \( -\frac{1}{k}\cos(kx) \), so for \( \sin(5x) \), \( k = 5 \). Thus, \( \int \frac{1}{2} \sin(5x) \ dx = -\frac{1}{2} \cdot \frac{1}{5} \cos(5x) = -\frac{1}{10}\cos(5x) \).
04

Integrate the Second Term

Now integrate \( \int \frac{1}{2} \sin(x) \ dx \). The integral of \( \sin(x) \) is \( -\cos(x) \). Therefore, \( \int \frac{1}{2} \sin(x) \ dx = -\frac{1}{2} \cos(x) \).
05

Combine the Results

Combine the results from Steps 3 and 4: \( \int \sin(2x) \cos(3x) \ dx = -\frac{1}{10}\cos(5x) + \frac{1}{2}\cos(x) + C \), where \( C \) is the constant of integration.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Product-to-Sum Formulas
The product-to-sum formulas are handy tools in trigonometry that help simplify expressions involving products of trigonometric functions. In this problem, we use it to convert a product of sine and cosine into a sum. This makes integration straightforward.

The specific formula used is:
  • \( \sin A \cos B = \frac{1}{2} [\sin(A+B) + \sin(A-B)] \).
By applying this formula to \( \sin 2x \cos 3x \), we can express it as:
  • \( \frac{1}{2} [\sin(5x) + \sin(-x)] \).
Recognizing that \( \sin(-x) = -\sin(x) \) allows us to simplify further to:
  • \( \frac{1}{2} [\sin(5x) - \sin(x)] \).
This transformation is crucial because it changes a seemingly complex product into a simple subtraction of sines, paving the way for easier integration.
Trigonometric Integrals
In calculus, integrating trigonometric functions is a common task. For the expression \( \frac{1}{2} [\sin(5x) - \sin(x)] \), we need to integrate each term separately.

Let's break it down:
  • First, consider \( \int \frac{1}{2} \sin(5x) \ dx \).
Here, the integral of \( \sin(kx) \) is \(-\frac{1}{k} \cos(kx)\). So for \( \sin(5x) \) where \( k = 5 \), we find:
  • \(-\frac{1}{10} \cos(5x)\).
Next, for \( \int \frac{1}{2} \sin(x) \ dx \):
  • The integral of \( \sin(x) \) is \(-\cos(x)\), resulting in \(-\frac{1}{2} \cos(x)\).
By integrating these components separately, we effectively simplify a complex trigonometric expression into manageable terms.
Constant of Integration
Every indefinite integral comes with a constant of integration, represented as \( C \). This constant accounts for the fact that integration is the reverse process of differentiation, where any constant term disappears during differentiation.

Without specifying \( C \), the solution could miss additional components. In the context of our problem:
  • The complete integral is \(-\frac{1}{10} \cos(5x) + \frac{1}{2} \cos(x) + C\).
Including \( C \) is crucial because it indicates the family of solutions possible, reflecting any vertical shifts in the graph of the function resulting from the integration.

It essentially ensures that whether you're evaluating or applying these results, you account for all potential options that the indefinite integral could imply.

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Most popular questions from this chapter

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