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Chapter 7: Problem 28

Evaluate the following integrals. $$\int e^{-2 \theta} \sin 6 \theta d \theta$$

Short Answer

Expert verified
Question: Determine the integral of the function \(e^{-2 \theta} \sin 6 \theta\) with respect to \(\theta\). Answer: \(\int e^{-2 \theta} \sin 6 \theta d \theta = -\frac{3}{4}e^{-2\theta}\cos6\theta + \frac{1}{8}e^{-2\theta}\sin6\theta + C\), where C is the integration constant.

Step by step solution

01

Choose u and dv

Select the functions according to the integration by parts rule. Choose: $$u = e^{-2\theta}$$ $$dv = \sin 6\theta d\theta$$
02

Determine du and v

Differentiate u to find du, and integrate dv to find v: $$du = -2e^{-2\theta} d\theta$$ $$v = -\frac{1}{6}\cos 6\theta$$
03

Apply integration by parts

Apply the integration by parts formula: $$\int e^{-2 \theta} \sin 6 \theta d \theta = \left(-\frac{1}{6}e^{-2\theta}\cos6\theta\right) - \int \left(-\frac{1}{6}\cos6\theta\right) (-2e^{-2\theta}d\theta)$$ Simplify the expression: $$\int e^{-2 \theta} \sin 6 \theta d \theta = \left(-\frac{1}{6}e^{-2\theta}\cos6\theta\right) + \frac{1}{3} \int e^{-2\theta} \cos6\theta d\theta$$
04

Apply integration by parts again

Now, apply integration by parts again for the integral on the right hand side with \(u=e^{-2\theta}\), and \(dv=\cos6\theta d\theta\). We obtain: $$du = -2e^{-2\theta} d\theta$$ $$v = \frac{1}{6}\sin6\theta$$ Plugging the values into the integration by parts formula, $$\frac{1}{3} \int e^{-2\theta} \cos6\theta d\theta = \frac{1}{3}\left(\frac{1}{6}e^{-2\theta}\sin6\theta - \int \frac{1}{6}\sin6\theta(-2e^{-2\theta}d\theta) \right)$$
05

Simplify and solve the integral

Simplify the expression and solve the final integral: $$\frac{1}{3} \int e^{-2\theta} \cos6\theta d\theta = \frac{1}{18}e^{-2\theta}\sin6\theta + \frac{1}{3} \int \frac{1}{3}e^{-2\theta} \sin6\theta d\theta$$ Now plug the expression for \(\frac{1}{3} \int e^{-2\theta} \cos6\theta d\theta\) back into the original integral: $$\int e^{-2 \theta} \sin 6 \theta d \theta = \left(-\frac{1}{6}e^{-2\theta}\cos6\theta\right) + \frac{1}{18}e^{-2\theta}\sin6\theta + \frac{1}{9} \int e^{-2\theta} \sin6\theta d\theta$$ To solve the equation, move the integral term to the left side: $$\frac{8}{9} \int e^{-2 \theta} \sin 6 \theta d \theta = \left(-\frac{1}{6}e^{-2\theta}\cos6\theta + \frac{1}{18}e^{-2\theta}\sin6\theta\right) $$ Now, divide by \(\frac{8}{9}\) to find the final solution: $$\int e^{-2 \theta} \sin 6 \theta d \theta = -\frac{3}{4}e^{-2\theta}\cos6\theta + \frac{1}{8}e^{-2\theta}\sin6\theta + C$$ Where C is the integration constant.

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Key Concepts

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

Definite Integrals
When it comes to definite integrals, understanding the concept is crucial. A definite integral is an integral that has specific upper and lower limits. These limits indicate the range over which the integration is performed. Unlike indefinite integrals, which result in a general function plus a constant, definite integrals yield a numerical value that represents the area under a curve between two points along the x-axis.

In solving definite integrals:
  • Identifying the limits of integration is the first step.
  • Evaluate the antiderivative at these limits and subtract to find the area.
It's important to note that definite integrals have numerous applications in calculating areas, solving physics problems, and more. In our specific example, the problem requires understanding and applying the integration technique of integration by parts, but it's also essential to recognize whether the integration is definite or indefinite and apply the matching rules accordingly.
Differentiation
Differentiation plays a significant role when we discuss integration, especially techniques like integration by parts. Differentiation is the process of finding the derivative of a function, which provides the rate at which a quantity changes.
To solve integrals:
  • One must understand how to differentiate functions effectively.
  • To differentiate, apply rules such as the product, quotient, and chain rules.
In integration by parts, a technique often used for integrating products of functions, the formula \( \int u\, dv = uv - \int v\, du \) heavily relies on differentiation. Here, you need to choose \( u \) and \( dv \) so that when you find \( du \), the derivative, and \( v \), the antiderivative, they simplify the integral. Mastery of differentiation hence aids in transforming complex integrals into manageable parts.
Transcendental Functions
Transcendental functions are a bit different from the polynomial or algebraic functions. They include exponential functions, logarithms, trigonometric functions, and their inverses. In our exercise, an example is the exponential function \( e^{-2\theta} \) and the trigonometric function \( \sin 6\theta \).

In mathematical problems:
  • Transcendental functions often require special techniques, like integration by parts, to integrate effectively.
  • They do not typically conform to simple algebraic formulas, offering a range of complexities.
With these functions, understanding their derivatives and antiderivatives is key. For example, the derivative of \( e^{x} \) is \( e^{x} \), and for sine and cosine functions, it's \( \cos x \) and \(-\sin x \) respectively. Recognizing these can simplify the process of solving integrals involving such functions, which is evident in our given solution steps.

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