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Chapter 8: Problem 22

Use the table of integrals at the back of the book to evaluate the integrals in Exercises \(1-26 .\) $$ \int \sin 2 x \cos 3 x d x $$

Short Answer

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

Step by step solution

01

Use the product-to-sum identities

The integral \( \int \sin 2x \cos 3x \, dx \) can be simplified by using the product-to-sum identities. These identities are helpful for converting products of sine and cosine into sums, making them easier to integrate. Specifically, use the identity: \[ \sin A \cos B = \frac{1}{2} \left( \sin(A + B) + \sin(A - B) \right) \]Let \( A = 2x \) and \( B = 3x \), so:\[ \sin 2x \cos 3x = \frac{1}{2} (\sin(5x) + \sin(-x)) \]
02

Simplify the expression

Substitute the expression from Step 1 into the integral:\[ \int \sin 2x \cos 3x \, dx = \int \frac{1}{2} (\sin(5x) + \sin(-x)) \, dx \]Separate the integrals:\[ = \frac{1}{2} \left( \int \sin(5x) \, dx + \int \sin(-x) \, dx \right) \]
03

Integrate each term

Begin by integrating each term individually:For \( \int \sin(5x) \, dx \):Use the integral formula \( \int \sin(kx) \, dx = -\frac{1}{k} \cos(kx) + C \).Thus:\[ \int \sin(5x) \, dx = -\frac{1}{5} \cos(5x) + C_1 \]For \( \int \sin(-x) \, dx \):Use the identity \( \sin(-x) = -\sin(x) \) and the same integral formula.So:\[ \int \sin(-x) \, dx = \int -\sin(x) \, dx = \cos(x) + C_2 \]
04

Combine the results

Substitute the results from each integral back into the expression:\[ \frac{1}{2} \left( -\frac{1}{5} \cos(5x) + \cos(x) \right) + C \] Simplify:\[ = -\frac{1}{10} \cos(5x) + \frac{1}{2} \cos(x) + C \] where \( C \) is the constant of integration, which accounts for both \( C_1 \) and \( C_2 \).

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Key Concepts

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

Trigonometric Integrals
Trigonometric integrals involve integrating functions of trigonometric expressions, like sine, cosine, or tangent. These integrals often appear in calculus problems and can be tricky due to the oscillatory nature of these functions. When working with trigonometric integrals, you might encounter products of different trigonometric functions. To solve these, you sometimes have to resort to trigonometric identities or substitution methods.
  • In problems like \( \int \sin 2x \cos 3x \, dx \), rather than integrating the complex product directly, simplification techniques are used to make the process easier.
  • Understanding the basic properties of sine and cosine functions, such as their periodicity and symmetry, will help you with your integration work.
  • Mastery of trigonometric identities is key for solving integrals efficiently and swiftly.
Product-to-Sum Identities
The product-to-sum identities are invaluable tools in trigonometric calculus. They allow us to transform products of trigonometric functions into sums that are often easier to integrate or differentiate. For instance, integrating a product like \( \sin 2x \cos 3x \) directly can be quite challenging.
  • By using the identity \( \sin A \cos B = \frac{1}{2} (\sin(A + B) + \sin(A - B)) \), the product of \( \sin 2x \) and \( \cos 3x \) converts into \( \frac{1}{2} (\sin 5x + \sin(-x)) \).
  • This transformation simplifies the integral into a more approachable form.
  • Remember, these identities are not just theoretical; they have practical applications, making complex integrals more straightforward.
Integration Techniques
Integration techniques cover various methods and strategies used to evaluate integrals, especially those involving complex functions. In the context of our example, breaking down the integral using trigonometric identities is one such technique. Here's a closer look at how these techniques come into play:
  • By transforming \( \int \sin 2x \cos 3x \, dx \) into sums using product-to-sum identities, the integration becomes more manageable.
  • Each resulting term, \( \sin 5x \) and \( \sin(-x) \), can be tackled with standard integration formulas.
  • The solution involves recognizing these simple forms and applying the formula \( \int \sin(kx) \, dx = -\frac{1}{k} \cos(kx) + C \).
  • Knowing when and how to apply these techniques can save a significant amount of time and effort.
Constant of Integration
The constant of integration, often represented as \( C \), is a fundamental concept in solving indefinite integrals. After integrating a function, the result isn't just a single antiderivative but an entire family of functions. This is where the constant of integration comes in:
  • When you perform an indefinite integration, \( C \) signifies an unknown constant that accounts for all possible vertical shifts of the function.
  • In our solution, \( C \) combines terms like \( C_1 \) and \( C_2 \) from the integration of separate parts, ensuring each possible antiderivative is covered.
  • The constant of integration reminds us that without additional conditions, like boundary values or initial conditions, we cannot solve for a unique solution.
  • This concept is crucial not only in pure mathematics but also in physics and engineering, where it can determine boundary conditions for real-world problems.

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Most popular questions from this chapter

Find the value of \(c\) so that \(f(x)=c \sqrt{x}(1-x)\) is a probability density function for the random variable \(X\) over \([0,1],\) and find the probability \(P(0.25 \leq X \leq 0.5).\)

Find the values of \(p\) for which each integral converges. \begin{equation} \text { a. }\int_{1}^{2} \frac{d x}{x(\ln x)^{p}} \quad \text { b. } \int_{2}^{\infty} \frac{d x}{x(\ln x)^{p}} \end{equation}

Lifetime of a tire Assume the random variable \(L\) in Example 2\(f\) is normally distributed with mean \(\mu=22,000\) miles and \(\sigma=4,000\) miles. a. In a batch of 4000 tires, how many can be expected to last for at least \(18,000\) miles? b. What is the minimum number of miles you would expect to find as the lifetime for 90\(\%\) of the tires?

Usable values of the sine-integral function The sine-integral function, $$\operatorname{Si}(x)=\int_{0}^{x} \frac{\sin t}{t} d t$$ is one of the many functions in engineering whose formulas cannot be simplified. There is no elementary formula for the antiderivative of \((\sin t) / t .\) The values of \(S \mathrm{Si}(x),\) however, are readily estimated by numerical integration. Although the notation does not show it explicitly, the function being integrated is $$f(t)=\left\\{\begin{array}{cl}{\frac{\sin t}{t},} & {t \neq 0} \\ {1,} & {t=0}\end{array}\right.$$ the continuous extension of \((\sin t) / t\) to the interval \([0, x]\) . The function has derivatives of all orders at every point of its domain. Its graph is smooth, and you can expect good results from Simpson's Rule. a. Use the fact that \(\left|f^{(4)}\right| \leq 1\) on \([0, \pi / 2]\) to give an upper bound for the error that will occur if $$\mathrm{Si}\left(\frac{\pi}{2}\right)=\int_{0}^{\pi / 2} \frac{\sin t}{t} d t$$ is estimated by Simpson's Rule with \(n=4\) b. Estimate \(\operatorname{Si}(\pi / 2)\) by Simpson's Rule with \(n=4\) . c. Express the error bound you found in part (a) as a percentage of the value you found in part (b).

Prove that the sum \(T\) in the Trapezoidal Rule for \(\int_{a}^{b} f(x) d x\) is a Riemann sum for \(f\) continuous on \([a, b] .\) (Hint: Use the Intermediate Value Theorem to show the existence of \(c_{k}\) in the sub interval \(\left[x_{k-1}, x_{k}\right]\) satisfying \(f\left(c_{k}\right)=\left(f\left(x_{k-1}\right)+f\left(x_{k}\right)\right) / 2 . )\)
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