/*! 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 54 Evaluate the integrals in Exerci... [FREE SOLUTION] | 91Ó°ÊÓ

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

Short Answer

Expert verified
The integral evaluates to \(\frac{1}{2} \arcsin(\sqrt{3/2}) + \frac{\sqrt{3}}{8}\).

Step by step solution

01

Identify substitution

We start by identifying a substitution that simplifies the integral. The term \(1 - y^2\) in the integrand suggests substituting \(y = \sin(\theta)\). This allows us to use the identity \(1 - \sin^2(\theta) = \cos^2(\theta)\).
02

Apply the substitution

By the substitution \(y = \sin(\theta)\), the differential \(dy = \cos(\theta) d\theta\). The limits of integration change as follows: when \(y = 0\), \(\theta = 0\), and when \(y = \sqrt{3/2}\), \(\theta = \arcsin(\sqrt{3/2})\). The integral becomes:\[\int_{0}^{\arcsin(\sqrt{3/2})} \frac{\cos(\theta) d\theta}{(1 - \sin^2(\theta))^{5/2}} = \int_{0}^{\arcsin(\sqrt{3/2})} \cos^{2} (\theta) d\theta.\]
03

Use power-reduction formula

The expression \(\cos^2(\theta)\) can be further simplified using the power-reduction identity: \(\cos^2(\theta) = \frac{1 + \cos(2\theta)}{2}\). Thus, the integral becomes:\[\int_{0}^{\arcsin(\sqrt{3/2})} \frac{1 + \cos(2\theta)}{2} d\theta.\]
04

Evaluate the simplified integral

The integral can now be split into two simpler integrals:\[\frac{1}{2} \int_{0}^{\arcsin(\sqrt{3/2})} d\theta + \frac{1}{2} \int_{0}^{\arcsin(\sqrt{3/2})} \cos(2\theta) d\theta.\]The first integral is straightforward:\[\frac{1}{2} \left[ \theta \right]_{0}^{\arcsin(\sqrt{3/2})} = \frac{1}{2} \arcsin(\sqrt{3/2}).\]
05

Evaluate the second integral

For the second integral, perform a substitution \(u = 2\theta\), so \(du = 2 d\theta\), or \(d\theta = \frac{1}{2} du\). Change the limits accordingly, and the integral becomes:\[\frac{1}{4} \int_{0}^{2 \arcsin(\sqrt{3/2})} \cos(u) du.\]Integrate:\[\frac{1}{4} \left[ \sin(u) \right]_{0}^{2 \arcsin(\sqrt{3/2})} = \frac{1}{4} [ \sin(2 \arcsin(\sqrt{3/2})) - \sin(0) ].\]
06

Simplify and conclude

The term \(\sin(2 \arcsin(\sqrt{3/2}))\) can be simplified using the double angle formula: \(\sin(2\theta) = 2\sin(\theta)\cos(\theta)\). Knowing that \(\sin(\theta) = \sqrt{3/2}\), \(\cos(\theta) = \sqrt{1 - (\sqrt{3/2})^2} = 1/2\), the expression simplifies to:\[2 \cdot \frac{\sqrt{3}}{2} \cdot \frac{1}{2} = \sqrt{3}/2.\]Thus, the integral simplifies to:\[\frac{1}{2} \arcsin(\sqrt{3/2}) + \frac{\sqrt{3}}{8}.\]

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Key Concepts

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

Substitution Method
The substitution method is a powerful tool in calculus used to simplify complex integrals by transforming them into a more manageable form. In our given exercise, the integral involves the expression \(1 - y^2\), which hints at a trigonometric identity. By choosing \(y = \sin(\theta)\), we can leverage the identity \(1 - \sin^2(\theta) = \cos^2(\theta)\), simplifying our integrand. This transformation also necessitates finding the new differential, \(dy = \cos(\theta) \, d\theta\). This substitution converts the variable from \(y\) to \(\theta\), making the integration process simpler.

Additionally, the substitution method often involves transforming the limits of integration. For instance, as \(y\) varies from 0 to \(\sqrt{3/2}\), \(\theta\) varies from 0 to \(\arcsin(\sqrt{3/2})\). Aligning the limits ensures that the integral maintains its path, crucial for obtaining the correct evaluation. Remember, it's like changing the perspective of a problem to make it easier to handle.
Trigonometric Substitution
Trigonometric substitution is a technique used when faced with expressions involving square roots, like \(\sqrt{1-y^2}\) in this exercise. Such terms often hint at employing trigonometric identities. By substituting \(y = \sin(\theta)\), we exploit these identities to replace \(\sqrt{1-y^2}\) with a simpler \(\cos(\theta)\).

This approach simplifies the integral significantly — what starts as a complicated rational function turns into a manageable trigonometric function. It's like using a key to unlock a door, where the trigonometric identities are the key and the complex expression is the locked door. This method is particularly useful for integrals involving square roots of quadratic expressions, which frequently appear in calculations involving circles and spheres.
Reduction Formula
The reduction formula is an advanced technique in integration that relates the integral of a function to the integral of its derivative, recursively simplifying the problem. In our particular problem, once we perform the trigonometric substitution, we use a power-reduction identity on \(\cos^2(\theta)\), specifically \(\cos^2(\theta) = \frac{1 + \cos(2\theta)}{2}\).

With this formula application, the complex integral is broken down into simpler parts. Our integral \(\int \cos^2(\theta) \, d\theta\) becomes two separate integrals, making it much easier to evaluate. Reduction formulas are invaluable, especially for integrals where straightforward substitution does not simplify the expression sufficiently. They transform complex polynomials or power functions into a sequence of simpler integrals to be computed step-by-step.
Integration Techniques
Integration techniques are diverse strategies used to evaluate integrals and can be crucial for solving complex problems. In this exercise, we primarily used a combination of substitution, trigonometric identities, and reduction formulas.

Each technique serves a distinct purpose: substitution changes the variable to something more familiar or simpler. Trigonometric substitution takes advantage of trig identities to address issues involving square roots and powers. Finally, reduction formulae reduce the powers of functions iteratively to facilitate easier integration.

This blend of methods is crucial for a problem like this. Mastering these techniques is important, as each type of problem might require different approaches. It's like having a toolkit where each tool has its purpose; picking the right one depends on the nature of the integral at hand. Ultimately, familiarity with various integration techniques can dramatically ease the process of finding solutions to challenging integrals.

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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 \(p\) . $$\int_{0}^{e} x^{p} \ln x d x$$

\begin{equation} \begin{array}{l}{\text { Consider the infinite region in the first quadrant bounded by the }} \\ {\text { graphs of } y=\frac{1}{x^{2}}, y=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}

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

Quality control A manufacturer of generator shafts finds that it needs to add additional weight to its shafts in order to achieve proper static and dynamic balance. Based on experimental tests, the average weight it needs to add is \(\mu=35 \mathrm{gm}\) with \(\sigma=9 \mathrm{gm}\) . Assuming a normal distribution, from 1000 randomly selected shafts, how many would be expected to need an added weight in excess of 40 \(\mathrm{gm} ?\)

\begin{equation} \begin{array}{l}{\text { Exercises } 71-74 \text { are about the infinite region in the first quadrant }} \\ {\text { between the curve } y=e^{-x} \text { and the } x \text { -axis. }}\end{array} \end{equation} Find the centroid of the region.

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