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Chapter 9: Problem 21

Evaluate the integral. $$\int_{0}^{\pi / 6} \sin 2 x \cos 4 x d x$$

Short Answer

Expert verified
The integral evaluates to \( \frac{1}{24} \).

Step by step solution

01

Apply the Product-to-Sum Formula

To simplify the integral, use the product-to-sum identities. The identity relevant here is \[ \sin A \cos B = \frac{1}{2}(\sin(A+B) + \sin(A-B)) \].By identifying \( A = 2x \) and \( B = 4x \), we find:\[ \sin 2x \cos 4x = \frac{1}{2}(\sin(2x + 4x) + \sin(2x - 4x)) = \frac{1}{2}(\sin 6x - \sin 2x) \].The integral becomes:\[ \int_{0}^{\pi/6} \left( \frac{1}{2} \sin 6x - \frac{1}{2} \sin 2x \right) dx \].
02

Split the Integral

Next, separate the original integral into two integrals:\[ \frac{1}{2} \int_{0}^{\pi/6} \sin 6x \, dx - \frac{1}{2} \int_{0}^{\pi/6} \sin 2x \, dx \].
03

Integrate Each Term

Integrate \( \sin 6x \) and \( \sin 2x \) separately:- The integral of \( \sin 6x \) is: \[ \int \sin 6x \, dx = -\frac{1}{6} \cos 6x \].- The integral of \( \sin 2x \) is: \[ \int \sin 2x \, dx = -\frac{1}{2} \cos 2x \].Thus, the integrals become:\[ \frac{1}{2} \left[-\frac{1}{6} \cos 6x \right]_{0}^{\pi/6} - \frac{1}{2} \left[-\frac{1}{2} \cos 2x \right]_{0}^{\pi/6} \].
04

Evaluate the Definite Integrals

Evaluate each definite integral:- For \( \int_{0}^{\pi/6} \sin 6x \, dx \): \[ -\frac{1}{6} \cos 6(\pi/6) + \frac{1}{6} \cos 6(0) = -\frac{1}{6} \cos \pi + \frac{1}{6} \cos 0 \]. \( \cos \pi = -1 \) and \( \cos 0 = 1 \), hence this evaluates to: \[ \frac{1}{6} (1 + 1) = \frac{1}{3} \].- For \( \int_{0}^{\pi/6} \sin 2x \, dx \): \[ -\frac{1}{2} \cos 2(\pi/6) + \frac{1}{2} \cos 2(0) = -\frac{1}{2} \cos \frac{\pi}{3} + \frac{1}{2} \cos 0 \]. \( \cos \frac{\pi}{3} = \frac{1}{2} \) and \( \cos 0 = 1 \), so this evaluates to: \[ \frac{1}{2} \left( -\frac{1}{2} + 1 \right) = \frac{1}{2} \left( \frac{1}{2} \right) = \frac{1}{4} \].
05

Combine the Results

Combine the results from the evaluated integrals:\[ \frac{1}{2} \cdot \frac{1}{3} - \frac{1}{2} \cdot \frac{1}{4} = \frac{1}{6} - \frac{1}{8} \].Get a common denominator for addition or subtraction:\[ \frac{1}{6} = \frac{4}{24} \quad \text{and} \quad \frac{1}{8} = \frac{3}{24} \].Thus, \( \frac{4}{24} - \frac{3}{24} = \frac{1}{24} \).

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Key Concepts

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

Product-to-Sum Identities
In calculus, dealing with trigonometric integrals often requires simplification techniques. The product-to-sum identities help in transforming products of trigonometric functions into sums or differences, making integrals easier to calculate. These identities convert complex expressions into more straightforward ones, especially when dealing with integrals of products like \( \sin A \cos B \).

The relevant identity here is:
  • \( \sin A \cos B = \frac{1}{2}(\sin(A+B) + \sin(A-B)) \)
Using this, the expression \( \sin 2x \cos 4x \) is simplified to \( \frac{1}{2}(\sin 6x - \sin 2x) \). This simplification splits the integral into two parts, allowing us to integrate more efficiently with basic antiderivatives of sine functions. Transforming products to sums aids in easier application of integral properties.
Definite Integrals
Definite integrals are a fundamental concept in calculus. They evaluate the net area under a curve within a certain interval. For example, in the task of evaluating \( \int_{0}^{\pi/6} \sin 2x \cos 4x \ dx \), we calculate the area under the curve from \( x = 0 \) to \( x = \pi/6 \).

After transforming the integral with product-to-sum identities, we evaluated:
  • \( \frac{1}{2} \int_{0}^{\pi/6} \sin 6x \, dx \)
  • \( -\frac{1}{2} \int_{0}^{\pi/6} \sin 2x \, dx \)
The final step involves computing the definite integrals for these transformed functions. The limits of integration simplify the evaluation process by substituting boundary values into antiderivatives. Calculating each separately and then combining their results provides the overall area.
Trigonometric Integrals
Trigonometric integrals involve integrating expressions containing trigonometric functions of a variable. These integrals can be challenging due to the periodic nature and complexity of trig functions. However, understanding basic antiderivatives helps simplify the process. For example, the antiderivative of \( \sin kx \) is \( -\frac{1}{k} \cos kx \).

In our integration of \( \sin 6x \) and \( \sin 2x \):
  • \( \int \sin 6x \, dx = -\frac{1}{6} \cos 6x \)
  • \( \int \sin 2x \, dx = -\frac{1}{2} \cos 2x \)
Working with these basic integrals emphasizes the importance of recognizing patterns and using identities to simplify, enabling efficient computation and ensuring accuracy when calculating definite integrals.

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Most popular questions from this chapter

In each part, find the Laplace transform. (a) \(f(t)=t, s>0\) (b) \(f(t)=t^{2}, s>0\) (c) \(f(t)=\left\\{\begin{array}{ll}0, & t<3 \\ 1, & t \geq 3\end{array}, s>0\right.\)

(a) Make an appropriate \(u\) -substitution of the form \(u=x^{1 / n}\) \(u=(x+a)^{1 / n},\) or \(u=x^{n},\) and then use the Endpaper Integral Table to evaluate the integral. (b) If you have a CAS, use it to evaluate the integral, and then confirm that the result is equivalent to the one that you found in part (a). $$\int \frac{1}{x \sqrt{x^{3}-1}} d x$$

(a) Make an appropriate \(u\)-substitution of the form \(u=x^{1 / n}\) \(u=(x+a)^{1 / n},\) or \(u=x^{n},\) and then use the Endpaper Integral Table to evaluate the integral. (b) If you have a CAS, use it to evaluate the integral, and then confirm that the result is equivalent to the one that you found in part (a). $$\int e^{\sqrt{x}} d x$$

Use any method to solve for \(x.\) $$\int_{1}^{x} \frac{1}{t \sqrt{2 t-1}} d t=1, x>\frac{1}{2}$$

Engineers want to construct a straight and level road 600 ft long and 75 ft wide by making a vertical cut through an intervening hill (see the accompanying figure). Heights of the hill above the centerline of the proposed road, as obtained at various points from a contour map of the region, are shown in the accompanying table. To estimate the construction costs, the engineers need to know the volume of earth that must be removed. Approximate this volume, rounded to the nearest cubic foot. [Hint: First, set up an integral for the cross-sectional area of the cut along the centerline of the road, then assume that the height of the hill does not vary between the centerline and edges of the road. \(]\) Figure cannot copy $$\begin{array}{cc} \hline \begin{array}{c} \text { HORIZONTAL } \\ \text { DISTANCE } x(\mathrm{ft}) \end{array} & \begin{array}{c} \text { HEIGHT } \\ h \text { (ft) } \end{array} \\ \hline 0 & 0 \\ 100 & 7 \\ 200 & 16 \\ 300 & 24 \\ 400 & 25 \\ 500 & 16 \\ 600 & 0 \\ \hline \end{array}$$
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