/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 55 Evaluate the integrals in Exerci... [FREE SOLUTION] | 91Ó°ÊÓ

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Evaluate the integrals in Exercises \(51-56\) $$ \int \cos 3 x \cos 4 x d x $$

Short Answer

Expert verified
\( \frac{1}{14} \sin 7x + \frac{1}{2} \sin x + C \)

Step by step solution

01

Apply Product-to-Sum Formulas

We know that the product of two cosine functions can be expressed using the product-to-sum identity. Specifically, for \(\cos A \cos B\), the identity is: \[ \cos A \cos B = \frac{1}{2} \left[ \cos(A+B) + \cos(A-B) \right] \]Here, \(A = 3x\) and \(B = 4x\), so we substitute to get:\[ \cos 3x \cos 4x = \frac{1}{2} \left[ \cos(3x + 4x) + \cos(3x - 4x) \right] = \frac{1}{2} \left[ \cos 7x + \cos (-x) \right] \] Since \(\cos(-x) = \cos x\), this simplifies to:\[ \cos 3x \cos 4x = \frac{1}{2} \left[ \cos 7x + \cos x \right] \]
02

Transform the Integral

Using the simplified expression from Step 1, substitute it back into the integral:\[ \int \cos 3x \cos 4x \, dx = \int \frac{1}{2} \left( \cos 7x + \cos x \right) \, dx \] This can be split using linearity of integration:\[ = \frac{1}{2} \left( \int \cos 7x \, dx + \int \cos x \, dx \right) \]
03

Integrate Each Cosine Function

Now, we integrate each term separately. For \(\int \cos 7x \, dx\), let \(u = 7x\), then \(du = 7 \, dx\) or \(dx = \frac{1}{7} du\):\[ \int \cos 7x \, dx = \frac{1}{7} \int \cos u \, du = \frac{1}{7} \sin u + C = \frac{1}{7} \sin 7x + C \] For \(\int \cos x \, dx\):\[ \int \cos x \, dx = \sin x + C \]
04

Combine the Results

Substitute the evaluated integrals back into the expression from Step 2:\[ \frac{1}{2} \left( \frac{1}{7} \sin 7x + \sin x \right) + C' \] Simplify:\[ = \frac{1}{14} \sin 7x + \frac{1}{2} \sin x + C' \] 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 Identities
When dealing with the product of trigonometric functions, transforming them using product-to-sum identities can simplify integration. These identities turn products into sums or differences, making them easier to handle mathematically. In particular, the identity for the product of two cosine functions is very useful:
  • \(\cos A \cos B = \frac{1}{2} [\cos(A+B) + \cos(A-B)]\)
Applying this to our problem, where \(A = 3x\) and \(B = 4x\), we transform \(\cos 3x \cos 4x\) into a sum of cosines:
  • \(\cos 3x \cos 4x = \frac{1}{2} [\cos 7x + \cos x]\)
This transformation helps in breaking down a complex integral into more manageable parts by reducing the multiplication of cosines to a simpler addition process. Understanding and mastering these identities is crucial for simplifying many trigonometric integrations.
Product-to-sum identities not only simplify computations but also provide deeper insights into the relationships between trigonometric functions and their applications in calculus.
Cosine Function
The cosine function, \(\cos x\), is one of the most fundamental trigonometric functions. It defines the cosine of an angle \(x\), which is the adjacent side over the hypotenuse in a right triangle setting. Cosine functions repeat their values in a periodic manner, which makes them especially useful in describing oscillations, such as waves.
Its basic properties include:
  • Periodicity: Repeats every \(2\pi\).
  • Even Function: \(\cos(-x) = \cos x\).
  • Range: Values lie between -1 and 1.
In the context of integration, the cosine function often appears with altered arguments like \(7x\) or \(3x + 4x\), corresponding to transformations seen in product-to-sum identities.
Understanding how to integrate cosine directly, as in \(\int \cos kx \, dx\), involves knowing antiderivatives, which are essential skills in calculus.

Integrating Cosine Functions

The integral of a cosine function, \(\int \cos kx \, dx\), is typically:
  • \(\frac{1}{k} \sin kx + C\)
This integration rule is crucial for solving problems involving harmonic motion or wave analysis, where the cosine function models periodic phenomena.
Indefinite Integration
Indefinite integration is a core concept in calculus that represents the reverse process of differentiation. It involves finding a function, called the antiderivative, whose derivative is the given function. The result of an indefinite integral includes a constant of integration, \(C\), because differentiation of a constant yields zero.
In our specific problem:
  • We're integrating \(\cos 3x \cos 4x\) after applying a product-to-sum identity.
This process demonstrates how indefinite integration requires breaking down complex expressions into smaller parts. By expressing \(\cos 3x \cos 4x\) using the identity:
  • \(\cos 3x \cos 4x = \frac{1}{2} [\cos 7x + \cos x]\)
we can separately integrate each term:
  • \(\int \cos 7x \, dx = \frac{1}{7} \sin 7x + C\)
  • \(\int \cos x \, dx = \sin x + C\)
Finally, these are combined to form the complete indefinite integrals solution. Integrating in this manner emphasizes the importance of recognizing patterns and identities that simplify calculations.

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Most popular questions from this chapter

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 \(\operatorname{Si}(x),\) however, are readily estimated by numerical integration. Although the notation does not show it explicitly, the function being integrated is $$ f(t)=\left\\{\begin{array}{cc}{\frac{\sin t}{t},} & {t \neq 0} \\ {1,} & {t=0}\end{array}\right. $$ 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 $$ \operatorname{Si}\left(\frac{\pi}{2}\right)=\int_{0}^{\pi / 2} \frac{\sin t}{t} d t $$ 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).

Use integration, the Direct Comparison Test, or the Limit Comparison Test to test the integrals for convergence. If more than one method applies, use whatever method you prefer. $$\int_{0}^{\pi} \frac{d t}{\sqrt{t}+\sin t}$$

Suppose you toss a fair coin \(n\) times and record the number of heads that land. Assume that \(n\) is large and approximate the discrete random variable \(X\) with a continuous random variable that is normally distributed with \(\mu=n / 2\) and \(\sigma=\sqrt{n} / 2 .\) If \(n=400\) find the given probabilities. $$ \begin{array}{ll}{\text { a. } P(190 \leq X<210)} & {\text { b. } P(X<170)} \\\ {\text { c. } P(X>220)} & {\text { d. } P(X=300)}\end{array} $$

Find the area of the region that lies between the curves \(y=\sec x\) and \(y=\tan x\) from \(x=0\) to \(x=\pi / 2\) .

Use integration, the Direct Comparison Test, or the Limit Comparison Test to test the integrals for convergence. If more than one method applies, use whatever method you prefer. $$\int_{e^{e}}^{\infty} \ln (\ln x) d x$$

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