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Chapter 8: Problem 47

Evaluate the integrals. Some integrals do not require integration by parts. $$ \int_{0}^{\pi / 2} \theta^{2} \sin 2 \theta d \theta $$

Short Answer

Expert verified
The evaluated integral is \( \frac{\pi^2 - 4}{8} \).

Step by step solution

01

Identify the Type of Integral

The integral given is \( \int_{0}^{\pi / 2} \theta^{2} \sin 2\theta \, d\theta \). This suggests the use of integration by parts because it involves a product of functions: a polynomial function \( \theta^2 \) and a trigonometric function \( \sin 2\theta \).
02

Choose Functions for Integration by Parts

Recall the integration by parts formula: \[ \int u \, dv = uv - \int v \, du \]. Let \( u = \theta^2 \), which makes \( du = 2\theta \, d\theta \), and choose \( dv = \sin 2\theta \, d\theta \). Integrating \( dv \), we get \( v = -\frac{1}{2}\cos 2\theta \).
03

Apply Integration by Parts Formula

Substitute \( u \), \( du \), \( v \), and \( dv \) into the integration by parts formula: \[ \int \theta^2 \sin 2\theta \, d\theta = \left[-\frac{1}{2} \theta^2 \cos 2\theta \right]_0^{\pi/2} + \frac{1}{2} \int_0^{\pi/2} \cos 2\theta \, (2\theta \, d\theta) \].
04

Evaluate Boundary Terms

Calculate the boundary terms \( \left[-\frac{1}{2} \theta^2 \cos 2\theta \right]_0^{\pi/2} \): At \( \theta = \pi/2 \), the term evaluates to \( -\frac{1}{2} \left( \frac{\pi}{2} \right)^2 \cos\pi = \frac{\pi^2}{8} \). At \( \theta = 0 \), the term is \( 0 \).Thus, the boundary evaluation is \( \frac{\pi^2}{8} \).
05

Simplify the New Integral

Now simplify \( \int_0^{\pi/2} 2\theta \cos 2\theta \). This requires another integration by parts:Let \( u = 2\theta \), so \( du = 2 \, d\theta \).Let \( dv = \cos 2\theta \, d\theta \), so \( v = \frac{1}{2} \sin 2\theta \).
06

Second Integration by Parts

Plug these into integration by parts: \[ \int 2\theta \cos 2\theta \, d\theta = \left[\theta \sin 2\theta \right]_0^{\pi/2} - \int \sin 2\theta \, d\theta \]. Evaluate the boundary and subsequent integral.
07

Evaluate Second Boundary Terms and New Integral

Evaluate boundary terms \( \left[\theta \sin 2\theta \right]_0^{\pi/2} \): At \( \theta = \pi/2 \), get \( \left( \frac{\pi}{2} \right)(0) = 0 \). At \( \theta = 0 \), result is \( 0 \). Resulting essential integral: \[ -\int_0^{\pi/2} \sin 2\theta \, d\theta = \left[-\frac{1}{2}\cos 2\theta \right]_0^{\pi/2} = \frac{1}{2} \].
08

Combine Results and Final Evaluation

Combine evaluated components: Boundary evaluations yield \( \frac{\pi^2}{8} \) for the first integration and \( \frac{1}{2} \) from integration over \( \cos 2\theta \).Hence, the integral evaluation is \[ \frac{\pi^2}{8} - \frac{1}{2} = \frac{\pi^2}{8} - \frac{4}{8} = \frac{\pi^2 - 4}{8} \].

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Key Concepts

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

Trigonometric Integrals
Trigonometric integrals involve expressions where trigonometric functions like sine and cosine appear inside the integral. They often require different techniques to solve based on their form.

In the given exercise, the integral has a trigonometric function, \( \sin 2\theta \), which plays a crucial part in the evaluation using integration by parts. The integration by parts method is particularly useful here because it allows breaking down the product of the polynomial function and the trigonometric function. Understanding trigonometric identities, such as how \( \sin 2\theta \) can be related to cosine, equips you with the skills to simplify and evaluate such integrals efficiently.

Trigonometric integrals are commonly encountered, and being familiar with trigonometric identities can greatly ease the process of finding their solutions.
Definite Integrals
Definite integrals are used to calculate the area under a curve between two specified points on the x-axis. They have specific upper and lower bounds, which makes it possible to find an exact value for the integral, unlike indefinite integrals, which represent a family of functions.

For the integral \( \int_{0}^{\pi / 2} \theta^{2} \sin 2\theta \, d\theta \), the bounds are from \( 0 \) to \( \frac{\pi}{2} \). These bounds were crucial when calculating the values at the endpoints while using the integration by parts technique. After deriving the antiderivative, the key step is to substitute the upper and lower limits to determine the net area under the curve.

It's essential to understand that in definite integrals, the limits provide the specific interval over which the area is calculated, ensuring you get precise and quantified results.
Integration Techniques
Integration techniques are methods used to solve various types of integrals. One of the most powerful techniques is integration by parts. It鈥檚 particularly useful when dealing with products of functions, such as the polynomial-trigonometric product in this problem.

In the integration by parts, you choose part of your integral to differentiate (\( u \)) and part to integrate (\( dv \)), using the formula \( \int u \, dv = uv - \int v \, du \). This approach transforms the integral into a simpler form or into a type that鈥檚 more easily solvable. In this problem, integration by parts was applied twice, as the resulting integral after the first pass still required simplification.

Here are some key tips for using integration by parts successfully:
  • Choose \( u \) and \( dv \) wisely to simplify your work (usually let \( u \) be a polynomial and \( dv \) a trigonometric or exponential function).
  • Be prepared to use the technique more than once if needed.
  • Don't forget to apply and evaluate the limits if it's a definite integral.
Mastering integration techniques comes with practice and boosts your ability to solve a wide range of integral problems effectively.

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Most popular questions from this chapter

In Exercises \(87-90,\) use a CAS to explore the integrals for various values of \(p\) (include noninteger values). For what values of \(p\) does the integral converge? What is the value of the integral when it does converge? Plot the integrand for various values of \(p\) . $$\int_{e}^{\infty} x^{p} \ln x d x$$

Evaluate the integrals in Exercises \(51-56\) by making a substitution (possibly trigonometric) and then applying a reduction formula. $$ \int_{0}^{\sqrt{3 / 2}} \frac{d y}{\left(1-y^{2}\right)^{5 / 2}} $$

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).

Use reduction formulas to evaluate the integrals in Exercises \(41-50 .\) $$ \int 8 \cot ^{4} t d t $$

\begin{equation} \begin{array}{l}{\text { Consider the infinite region in the first quadrant bounded by the }} \\ {\text { graphs of } y=\frac{1}{\sqrt{x}}, y=0, x=0, \text { and } x=1} \\ {\text { a. Find the area of the region. }} \\ {\text { b. Find the volume of the solid formed by revolving the region }} \\ {\text { (i) about the } x \text { -axis; (ii) about the } y \text { -axis. }}\end{array} \end{equation}
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