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Chapter 7: Problem 6

Evaluate the integral. $$\int \cos ^{3} a t d t$$

Short Answer

Expert verified
\(\int \cos^3(at) \, dt = \frac{3}{4a} \sin(at) + \frac{1}{12a} \sin(3at) + C\).

Step by step solution

01

Use Trigonometric Identity

The integral involves a power of cosine. Recall the trigonometric identity for powers of cosine: \[\cos^2(x) = \frac{1 + \cos(2x)}{2}\]This identity helps break down higher powers of cosine. First, express \(\cos^3(at)\) as \(\cos(at) \cdot \cos^2(at)\).
02

Substitute the Trigonometric Identity

Use the identity \(\cos^2(at) = \frac{1 + \cos(2at)}{2}\) in the expression:\[\cos^3(at) = \cos(at) \cdot \frac{1 + \cos(2at)}{2} = \frac{1}{2}\cos(at) + \frac{1}{2}\cos(at)\cos(2at)\].
03

Separate the Integral

Now separate the integral using this substitution:\[\int \cos^3(at) \, dt = \int \frac{1}{2}\cos(at) \, dt + \int \frac{1}{2}\cos(at)\cos(2at) \, dt\].
04

Evaluate the First Integral

The first integral is:\[\int \frac{1}{2} \cos(at) \, dt = \frac{1}{2} \cdot \frac{1}{a} \sin(at) + C_1\]where \(C_1\) is a constant of integration.
05

Use Product-to-Sum Formula for the Second Integral

For \(\int \frac{1}{2}\cos(at)\cos(2at) \, dt\), use the product-to-sum identity:\[\cos(A)\cos(B) = \frac{1}{2}(\cos(A+B) + \cos(A-B))\].Applying this identity gives:\[\cos(at)\cos(2at) = \frac{1}{2}(\cos(3at) + \cos(-at))\].This simplifies the integral to:\[\int \frac{1}{4}(\cos(3at) + \cos(-at)) \, dt\].
06

Evaluate the Transformed Second Integral

Evaluate \(\int \frac{1}{4}(\cos(3at) + \cos(-at)) \, dt\):\[\frac{1}{4}\int \cos(3at) \, dt + \frac{1}{4}\int \cos(at) \, dt\]\[= \frac{1}{4} \cdot \frac{1}{3a} \sin(3at) + \frac{1}{4} \cdot \frac{1}{a} \sin(at) + C_2\]where \(C_2\) is another constant of integration.
07

Combine the Results

Combining all parts of the solution:\[\int \cos^3(at) \, dt = \frac{1}{2a} \sin(at) + \frac{1}{12a} \sin(3at) + \frac{1}{4a} \sin(at) + C\]Simplify by combining like terms:\[= \left(\frac{1}{2a} + \frac{1}{4a}\right) \sin(at) + \frac{1}{12a} \sin(3at) + C\]\[= \frac{3}{4a} \sin(at) + \frac{1}{12a} \sin(3at) + C\]where \(C = C_1 + C_2\) is the overall constant of integration.

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Key Concepts

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

Trigonometric Identities
Trigonometric identities are very useful tools in calculus, particularly when dealing with integrals involving trigonometric functions. They help us simplify expressions and make it easier to work with them in various mathematical operations.
A core identity used in this exercise is \[\cos^2(x) = \frac{1 + \cos(2x)}{2}\] which allows us to express higher powers of cosine in a more manageable form.
Using this identity, we can decompose \(\cos^3(at)\) into a product of simpler functions:
  • First, express \(\cos^3(at)\) as \(\cos(at) \cdot \cos^2(at)\).
  • Substitute the identity \(\cos^2(at) = \frac{1 + \cos(2at)}{2}\) to simplify the expression into \(\frac{1}{2}\cos(at) + \frac{1}{2}\cos(at)\cos(2at)\).
Recognizing and applying these identities can turn complex expressions into sums of easier integrals, which are generally more straightforward to evaluate.
Integration Techniques
Integration techniques involve different methods used to evaluate integrals, especially when direct integration is not possible. Common techniques include substitution, integration by parts, and the use of trigonometric identities and transformations to simplify the integrand.
In the given exercise, after simplifying the trigonometric expression using identities, the integral was broken down into simpler parts:
  • The first integral, \(\int \frac{1}{2}\cos(at) \, dt\), is straightforward and leads to \(\frac{1}{2a} \sin(at) + C_1\), where \(C_1\) is a constant of integration.
  • The second integral, \(\int \frac{1}{2}\cos(at)\cos(2at) \, dt\), required additional steps using another powerful technique: the product-to-sum formulas.
Learning different integration techniques provides the flexibility to tackle a wide variety of problems efficiently. This problem exemplifies the need to break down complex tasks into simpler, manageable components.
Product-to-Sum Formulas
Product-to-sum formulas transform products of trigonometric functions into sums or differences, making them easier to integrate or differentiate.
The formula employed here is:\[\cos(A)\cos(B) = \frac{1}{2}(\cos(A+B) + \cos(A-B))\]By applying this formula, we managed to simplify the integral containing a product \(\cos(at)\cos(2at)\) into a sum of cosines:
  • Transformed into \(\frac{1}{4}(\cos(3at) + \cos(-at))\).
This transformation facilitated the integration of each part:
  • \(\int \frac{1}{4}\cos(3at) \, dt\) results in \(\frac{1}{12a} \sin(3at)\).
  • \(\int \frac{1}{4}\cos(at) \, dt\) simplifies to \(\frac{1}{4a} \sin(at)\).
  • These were then combined with an overall constant \(C_2\) of integration.
Using such formulas is not only a clever way to simplify complex trigonometric expressions, but it's also a strategy that can save time and reduce errors.

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Most popular questions from this chapter

In each part, try to evaluate the integral exactly with a CAS. If your result is not a simple numerical answer, then use the CAS to find a numerical approximation of the integral. (a) \(\int_{-\infty}^{+\infty} \frac{1}{x^{8}+x+1} d x\) (b) \(\int_{0}^{+\infty} \frac{1}{\sqrt{1+x^{3}}} d x\) (c) \(\int_{1}^{+\infty} \frac{\ln x}{e^{x}} d x\) (d) \(\int_{1}^{+\infty} \frac{\sin x}{x^{2}} d x\)

(a).Make \(u\) -substitution (5) to convert the integrand to a rational function of \(u\), and then evaluate the integral. (b).If you have a CAS, use it to evaluate the integral (no substitution), and then confirm that the result is equivalent to that in part (a). $$\int \frac{d x}{2+\sin x}$$

Suppose that the region between the \(x\) -axis and the curve \(y=e^{-x}\) for \(x \geq 0\) is revolved about the \(x\) -axis. (a) Find the volume of the solid that is generated. (b) Find the surface area of the solid.

In each part, confirm the result with a CAS. (a) \(\int_{0}^{+\infty} \frac{\sin x}{\sqrt{x}} d x=\sqrt{\frac{\pi}{2}}\) (b) \(\int_{-\infty}^{+\infty} e^{-x^{2}} d x=\sqrt{\pi}\) (c) \(\int_{0}^{1} \frac{\ln x}{1+x} d x=-\frac{\pi^{2}}{12}\)

The average speed, \(\bar{v},\) of the molecules of an ideal gas is given by $$\bar{v}=\frac{4}{\sqrt{\pi}}\left(\frac{M}{2 R T}\right)^{3 / 2} \int_{0}^{+\infty} v^{3} e^{-M v^{2} /(2 R T)} d v$$ and the root-mean-square speed, \(v_{\mathrm{rms}},\) by $$v_{\mathrm{rms}}^{2}=\frac{4}{\sqrt{\pi}}\left(\frac{M}{2 R T}\right)^{3 / 2} \int_{0}^{+\infty} v^{4} e^{-M v^{2} /(2 R T)} d v$$ where \(v\) is the molecular speed, \(T\) is the gas temperature, \(M\) is the molecular weight of the gas, and \(R\) is the gas constant. (a) Use a CAS to show that $$\int_{0}^{+\infty} x^{3} e^{-a^{2} x^{2}} d x=\frac{1}{2 a^{4}}, \quad a>0$$ and use this result to show that \(\bar{v}=\sqrt{8 R T /(\pi M)}\) (b) Use a CAS to show that $$\int_{0}^{+\infty} x^{4} e^{-a^{2} x^{2}} d x=\frac{3 \sqrt{\pi}}{8 a^{5}}, \quad a>0$$ and use this result to show that \(v_{\mathrm{rms}}=\sqrt{3 R T / M}\).
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