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Chapter 8: Problem 20

Evaluate the integrals. \(\int_{0}^{\pi} 8 \sin ^{4} y \cos ^{2} y d y\)

Short Answer

Expert verified
The integral evaluates to \( \frac{7\pi}{2} \).

Step by step solution

01

Use Trigonometric Identities

We begin by simplifying the integrand using trigonometric identities. Note that \( \sin^2 y = 1 - \cos^2 y \). We can write \( \sin^4 y = (\sin^2 y)^2 = (1 - \cos^2 y)^2 \) and substitute this into the integral.
02

Substitute and Simplify

Substitute \( \sin^4 y = (1 - \cos^2 y)^2 \) into the integral. The integral now becomes: \[ \int_{0}^{\pi} 8 (1 - \cos^2 y)^2 \cos^2 y \, dy. \] Simplify this to: \[ \int_{0}^{\pi} 8 (1 - 2\cos^2 y + \cos^4 y) \cos^2 y \, dy. \] This becomes: \[ \int_{0}^{\pi} 8 (\cos^2 y - 2\cos^4 y + \cos^6 y) \, dy. \]
03

Separate into Individual Integrals

Separate the integral into three simpler integrals: \[ 8 \left( \int_{0}^{\pi} \cos^2 y \, dy - 2 \int_{0}^{\pi} \cos^4 y \, dy + \int_{0}^{\pi} \cos^6 y \, dy \right). \] We will solve each of these integrals separately.
04

Evaluate \( \int_{0}^{\pi} \cos^2 y \, dy \)

Use the identity \( \cos^2 y = \frac{1 + \cos 2y}{2} \) to express the integral as \( \int_{0}^{\pi} \frac{1 + \cos 2y}{2} \, dy \). This simplifies to \( \frac{1}{2} \left( \int_{0}^{\pi} 1 \, dy + \int_{0}^{\pi} \cos 2y \, dy \right) \). The first integral is \( \int_{0}^{\pi} 1 \, dy = \pi \), and \( \int_{0}^{\pi} \cos 2y \, dy = 0 \). Hence, \( \int_{0}^{\pi} \cos^2 y \, dy = \frac{\pi}{2} \).
05

Evaluate \( \int_{0}^{\pi} \cos^4 y \, dy \)

Use the identity \( \cos^4 y = \left(\frac{1 + \cos 2y}{2}\right)^2 \), simplifying to \( \int_{0}^{\pi} \frac{1 + 2\cos 2y + \cos^2 2y}{4} \, dy \). Integrate term by term. The integrals of \(1\) and \(\cos 2y\) are as before. For \( \cos^2 2y \), reapply the identity to find \( \int_{0}^{\pi} \cos^2 2y \, dy = \frac{\pi}{2} \). Simplifying gives \( \int_{0}^{\pi} \cos^4 y \, dy = \frac{3\pi}{8} \).
06

Evaluate \( \int_{0}^{\pi} \cos^6 y \, dy \)

Express \( \cos^6 y \) as \( \left(\frac{1 + \cos 2y}{2}\right)^3 \) for further simplification. After simplification using similar steps as in the previous integrals, find the value as \( \frac{5\pi}{16} \).
07

Combine Results

Substitute back the results: \[ 8 \left( \frac{\pi}{2} - 2 \times \frac{3\pi}{8} + \frac{5\pi}{16} \right). \] Simplify the expression: \[ 8 \left( \frac{8\pi}{16} - \frac{6\pi}{16} + \frac{5\pi}{16} \right) = 8 \times \frac{7\pi}{16}. \] This results in \( \frac{56\pi}{16} = \frac{7\pi}{2} \).

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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 essential tools for simplifying complex trigonometric expressions and solving integrals involving trigonometric functions.
In this exercise, we utilize several trigonometric identities to transform the integral into a more manageable form.
  • The identity \( \sin^2 y = 1 - \cos^2 y \) helps convert powers of sine into cosine.
  • Using \( \cos^2 y = \frac{1 + \cos 2y}{2} \), the integral simplifies further, allowing us to break it down into simpler parts.
These identities simplify calculations and are commonly used in trigonometric integrals. Understanding how to apply them is key in integral calculus.
Integration by Substitution
Integration by substitution is a fundamental technique in calculus used to find integrals where direct integration is challenging.
It involves substituting a part of the integrand with a single variable, simplifying the expression. This method is particularly useful when dealing with complicated trigonometric expressions.
In this problem, we effectively applied substitution by replacing parts of the trigonometric expression with simpler forms using identities, making the integration process straightforward. Although explicit substitution isn't used, the principle of simplification via substitution plays a critical role.
Integral Calculus
Integral calculus deals with finding the antiderivatives of functions, calculating areas under curves, and solving accumulation problems.
In the exercise, we focus on evaluating definite integrals using integration techniques and trigonometric identities.
  • The concept of separating an integral into parts allows solving complex integrals as sums of simpler ones.
  • By solving each simple integral individually, we then combine the results to obtain the final answer.
Integral calculus is powerful for evaluating a wide range of problems involving continuous functions.
Definite Integrals
Definite integrals calculate the area under a curve over a specific interval. They provide exact values unlike indefinite integrals, which offer general solutions with unknown constants.
In this problem, we solved the integral from \(0\) to \(\pi\), which required careful evaluation of simpler integrals within the given interval.
  • Using definite integrals, we determine exact areas or quantities, such as the total accumulated area under a trigonometric curve over the specified bounds.
  • The results are precise, giving a clear and concrete value for the integral's outcome.
These integrals are key in both theoretical mathematics and practical applications, offering exact solutions in context.

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Most popular questions from this chapter

Use a CAS to evaluate the integrals. $$\int_{0}^{2 / \pi} \sin \frac{1}{x} d x$$

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

Centroid Find the centroid of the region cut from the first quadrant by the curve \(y=1 / \sqrt{x+1}\) and the line \(x=3 .\)

Use reduction formulas to evaluate the integrals in Exercises \(41-50 .\) $$ \int 2 \sec ^{3} \pi x d x $$

In Exercises 65 and \(66,\) use a CAS to perform the integrations. Evaluate the integrals $$ a.\int \frac{\ln x}{x^{2}} d x \quad \text { b. } \int \frac{\ln x}{x^{3}} d x \quad \text { c. } \int \frac{\ln x}{x^{4}} d x $$$$ \begin{array}{c}{\text { d. What pattern do you see? Predict the formula for }} \\ {\int \frac{\ln x}{x^{5}} d x} \\ {\text { and then see if you are correct by evaluating it with a CAS. }}\end{array} $$$$ \begin{array}{c}{\text { e. What is the formula for }} \\ {\int \frac{\ln x}{x^{n}} d x, \quad n \geq 2 ?}\end{array} $$ Check your answer using a CAS.
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