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Chapter 8: Problem 6

Evaluate the integrals. \(\int \cos ^{3} 4 x d x\)

Short Answer

Expert verified
The integral evaluates to \( \frac{1}{4}\sin(4x) - \frac{1}{12}\sin^3(4x) + C \).

Step by step solution

01

Express the Integral in Terms of Cosine and Sine

Start by rewriting the integral using trigonometric identities. Recall that \(\cos^3(4x)\) can be rewritten by using \(\cos^2(4x) = 1 - \sin^2(4x)\). Thus, \(\cos^3(4x) = \cos(4x) \cdot (1 - \sin^2(4x))\), which becomes \(\cos(4x) - \cos(4x)\sin^2(4x)\). This breaks the integral into two separate integrals: \(\int \cos(4x) \, dx - \int \cos(4x)\sin^2(4x) \, dx\).
02

Solve the First Integral

Integrate \(\int \cos(4x) \, dx\) directly. The antiderivative of \(\cos(4x)\) is \(\frac{1}{4} \sin(4x) + C\). So, the solution to this integral is \(\frac{1}{4} \sin(4x)\).
03

Use Substitution for the Second Integral

Apply substitution for \(\int \cos(4x) \sin^2(4x) \, dx\). Let \(u = \sin(4x)\), then \(du = 4\cos(4x)\,dx\) which implies \(dx = \frac{du}{4\cos(4x)}\). Substitute \(u\) back into the integral to write it as \(- \frac{1}{4} \int u^2 \, du\).
04

Integrate the Substitution Result

Integrate \(- \frac{1}{4} \int u^2 \, du\). The antiderivative of \(u^2\) is \(\frac{1}{3}u^3\). Thus, the integral becomes \(- \frac{1}{4} \cdot \frac{1}{3} u^3 = -\frac{1}{12} u^3\). Substitute back \(u = \sin(4x)\) to get \(-\frac{1}{12} (\sin(4x))^3\).
05

Combine Results for the Final Answer

Combine both results from Step 2 and Step 4. The expression becomes \(\frac{1}{4}\sin(4x) - \frac{1}{12}\sin^3(4x) + C\). This is the final evaluated integral.

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Key Concepts

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

Trigonometric Integrals
Trigonometric integrals are integrals of functions that contain trigonometric functions like sine, cosine, tangent, or their powers. For students tackling calculus, these types of integrals are very common.
They often require special techniques for solving, as straightforward integration is not always possible. When you encounter a trigonometric integral, you might need to:
  • Use trigonometric identities to simplify the integrand.
  • Apply substitution methods to make the integral more manageable.
  • Break the integral into multiple parts for easier integration.
Each of these techniques will help you transform an intricate trigonometric integral into simpler components. This will lead you to accurately calculate the antiderivative. Practice and familiarity with basic trigonometric identities is key here.
Trigonometric Identities
Trigonometric identities are equations that hold true for all angles and are used to simplify expressions and solve integrals. In the given problem, an identity simplifies the expression \(\cos^3(4x)\).
By using the identity \(\cos^2(4x) = 1 - \sin^2(4x)\), we rewrite \(\cos^3(4x)\) as \(\cos(4x)(1-\sin^2(4x))\).Useful trigonometric identities include:
  • Pythagorean Identities, like \(\sin^2(\theta) + \cos^2(\theta) = 1\)
  • Angle Sum and Difference Identities
  • Double Angle Identities
These identities are powerful tools. They simplify complex integrals, isolating terms or converting products into sums. Identifying which identity to use is crucial, as it can transform an integral into a solvable form.
Substitution Method
The substitution method, also known as u-substitution, is a strategy used to make integrals simpler to evaluate.
This method involves choosing a substitution that simplifies the original integral into a more familiar form.To apply substitution:
  • Select a part of the integral to substitute as \(u\)
  • Find \(du\), the differential of \(u\), and express \(dx\) in terms of \(du\)
  • Rewrite the integral in terms of \(u\)
  • Integrate in terms of \(u\)
  • Substitute back the original variable after integrating
In our problem, the substitution of \(u = \sin(4x)\) enabled us to transform the integral \(\int \cos(4x) \sin^2(4x) \, dx\) into a polynomial in \(u\). This conversion allows for straightforward integration.
Antiderivative
The antiderivative is the reverse process of differentiation. Finding the antiderivative means finding the original function given its derivative, usually expressed as an indefinite integral.
The final step in solving integrals involves determining this antiderivative (also called the integral).For example, the antiderivative of \(\cos(4x)\) is \(\frac{1}{4} \sin(4x) + C\) because the derivative of \(\frac{1}{4} \sin(4x)\) yields \(\cos(4x)\), illustrating this reverse process.Key points to consider:
  • Always add the constant of integration (\(C\)) after finding the antiderivative.
  • The process of integration can often involve recognizing derivatives of familiar functions.
  • Techniques like substitution or partial fractions may be required to find complex antiderivatives.
Understanding antiderivatives is central in calculus, as they allow us to reverse the differentiation process, effectively solving a wide range of real-world problems.

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Most popular questions from this chapter

In Exercises 65 and \(66,\) use a CAS to perform the integrations. Evaluate the integrals $$ (a)\int x \ln x d x \quad \text { b. } \int x^{2} \ln x d x \quad \text { c. } \int x^{3} \ln x d x $$ $$ \begin{array}{l}{\text { d. What pattern do you see? Predict the formula for } \int x^{4} \ln x d x} \\ {\text { and then see if you are correct by evaluating it with a CAS. }} \\ {\text { e. What is the formula for } \int x^{n} \ln x d x, n \geq 1 ? \text { Check your }} \\ {\text { answer using a CAS. }}\end{array} $$

Show that if \(f(x)\) is integrable on every interval of real numbers and \(a\) and \(b\) are real numbers with \(a < b,\) then \begin{equation} \begin{array}{l}{\text { a. } \int_{-\infty}^{a} f(x) d x \text { and } \int_{a}^{\infty} f(x) d x \text { both converge if and only if }} \\\ {\int_{-\infty}^{b} f(x) d x \text { and } \int_{b}^{\infty} f(x) d x \text { both converge. }} \\ {\text { b. } \int_{-\infty}^{a} f(x) d x+\int_{a}^{\infty} f(x) d x=\int_{-\infty}^{b} f(x) d x+\int_{b}^{\infty} f(x) d x} \\ {\quad \text { when the integrals involved converge. }}\end{array} \end{equation}

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. a. \(P(190 \leq X<210) \quad\) b. \(P(X<170)\) c. \(P(X>220) \quad\) d. \(P(X=300)\)

The infinite paint can or Gabriel's horn As Example 3 shows, the integral \(\int_{1}^{\infty}(d x / x)\) diverges. This means that the integral $$\int_{1}^{\infty} 2 \pi \frac{1}{x} \sqrt{1+\frac{1}{x^{4}}} d x$$ which measures the surface area of the solid of revolution traced out by revolving the curve \(y=1 / x, 1 \leq x,\) about the \(x\) -axis, diverges also. By comparing the two integrals, we see that, for every finite value \(b>1\) . $$\int_{1}^{b} 2 \pi \frac{1}{x} \sqrt{1+\frac{1}{x^{4}}} d x>2 \pi \int_{1}^{b} \frac{1}{x} d x$$ However, the integral $$\int_{1}^{\infty} \pi\left(\frac{1}{x}\right)^{2} d x$$ for the volume of the solid converges. \begin{equation} \begin{array}{l}{\text { a. Calculate it. }} \\ {\text { b. This solid of revolution is sometimes described as a can that }} \\ {\text { does not hold enough paint to cover its own interior. Think }} \\ {\text { about that for a moment. It is common sense that a finite }}\\\ {\text { amount of paint cannot cover an infinite surface. But if we fill }} \\\ {\text { the horn with paint (a finite amount), then we will have cov- }} \\\ {\text { ered an infinite surface. Explain the apparent contradiction. }}\end{array} \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?
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