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Determine whether the statement is true or false. Explain your answer. An integrand involving a radical of the form \(\sqrt{a^{2}-x^{2}}\) suggests the substitution \(x=a \sin \theta\).

Short Answer

Expert verified
True; the substitution \(x = a \sin \theta\) simplifies \(\sqrt{a^{2}-x^{2}}\) effectively.

Step by step solution

01

Understanding the given integrand

The given integrand involves a radical expression of the form \(\sqrt{a^{2}-x^{2}}\). Recognize this as a common structure in integrals and substitutions.
02

Identifying the suitable substitution

Recall that for integrands involving \(\sqrt{a^{2}-x^{2}}\), the trigonometric substitution \(x = a \sin \theta\) is typically used. This transformation leverages the Pythagorean identity \(1 - \sin^2 \theta = \cos^2 \theta\).
03

Substitute and simplify

Substitute \(x = a \sin \theta\) into the expression \(\sqrt{a^{2}-x^{2}}\). This results in \(\sqrt{a^{2} - (a \sin \theta)^{2}} = \sqrt{a^{2} - a^{2} \sin^{2} \theta} = \sqrt{a^{2}(1 - \sin^{2} \theta)} = a \cos \theta\).
04

Verify using the identity

Verify the simplification using the Pythagorean identity: \(\sin^2 \theta + \cos^2 \theta = 1\), so \(1 - \sin^2 \theta = \cos^2 \theta\). Thus, \(\sqrt{1 - \sin^2 \theta} = \cos \theta\), confirming that the substitution simplifies the expression correctly.
05

Determine the truth value of the statement

Since the substitution \(x = a \sin \theta\) effectively simplifies the integrand \(\sqrt{a^{2}-x^{2}}\) into \(a \cos \theta\), the statement is true. This is a standard trigonometric substitution for such integrals.

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Key Concepts

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

Integration Techniques
Integration techniques are methods used to find the integral of a function, and these often involve strategies for simplifying complex integrands. One popular technique is substitution, where we replace parts of the integrand with a new variable to make integration easier. This is particularly useful when the integrand includes radicals or trigonometric functions.

When faced with an integrand involving a square root, like \( \sqrt{a^2 - x^2} \), direct integration can be complicated. Instead, we use trigonometric substitution to transform and simplify the problem.
  • If the integrand includes \( \sqrt{a^2 - x^2} \), the substitution \( x = a \sin \theta \) can simplify the expression.
  • For \( \sqrt{a^2 + x^2} \), choose \( x = a \tan \theta \).
  • For \( \sqrt{x^2 - a^2} \), the substitution \( x = a \sec \theta \) is used.
With trigonometric substitution, the original integral is transformed into one involving trigonometric identities, which are often more straightforward to solve.
Pythagorean Identity
The Pythagorean identity is a cornerstone of trigonometry, expressing the fundamental relationship between the sine and cosine functions. It states: \( \sin^2 \theta + \cos^2 \theta = 1 \). This identity is used to simplify expressions and solve equations involving trigonometric functions.

When tackling an integrand with \( \sqrt{a^2 - x^2} \), the substitution \( x = a \sin \theta \) brings the identity into play. Here's how it works in the context of integration:
  • Using \( x = a \sin \theta \), we rewrite \( \sqrt{a^2 - x^2} \) as \( \sqrt{a^2(1 - \sin^2 \theta)} \).
  • Applying the Pythagorean identity, we see that \( 1 - \sin^2 \theta = \cos^2 \theta \).
  • Thus, \( \sqrt{a^2 \cos^2 \theta} = a \cos \theta \).
This simplification is crucial in converting a potentially difficult integral into a more tractable form that involves the cosine function rather than a square root expression. Understanding and utilizing the Pythagorean identity is key to successful trigonometric substitution.
Radical Expressions
Radical expressions involve roots, most commonly square roots, and present unique challenges in both algebra and calculus. When these expressions appear under integrals, determining an appropriate substitution can turn an intractable problem into a manageable one.

Consider the radical expression \( \sqrt{a^2 - x^2} \): this specific form immediately suggests a trigonometric substitution. By choosing \( x = a \sin \theta \), the expression transforms into \( a \cos \theta \) using the Pythagorean identity, thus removing the radical altogether.
  • This transformation leverages the relationship between trigonometric identities and radical expressions.
  • By removing the radical, the integration process is simplified significantly.
  • As a result, integrals that appear complex due to radical expressions often become straightforward with the right substitution.
Being able to identify and manipulate radical expressions effectively using trigonometric substitution is a valuable integration skill that helps simplify and solve complex integral problems.

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Most popular questions from this chapter

Medication can be administered to a patient using a variety of methods. For a given method, let \(c(t)\) denote the concentration of medication in the patient's bloodstream (measured in \(\mathrm{mg} / \mathrm{L}) t\) hours after the dose is given. The area under the curve \(c=c(t)\) over the time interval \([0,+\infty)\) indicates the "availability" of the medication for the patient's body. Determine which method provides the greater availability. Method \(1: c_{1}(t)=5\left(e^{-0.2 t}-e^{-t}\right)\) Method 2: \(c_{2}(t)=4\left(e^{-0.2 t}-e^{-3 t}\right)\)

(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). $$ \int \frac{d x}{\sin x+\tan x} $$

Let \(R\) be the region to the right of \(x=1\) that is bounded by the \(x\) -axis and the curve \(y=1 / x .\) When this region is revolved about the \(x\) -axis it generates a solid whose surface is known as Gabriel's Horn (for reasons that should be clear from the accompanying figure). Show that the solid has a finite volume but its surface has an infinite area. [Note: It has been suggested that if one could saturate the interior of the solid with paint and allow it to seep through to the surface, then one could paint an infinite surface with a finite amount of paint! What do you think?]

Use any method to solve for \(x\). $$ \int_{1}^{x} \frac{1}{t \sqrt{2 t-1}} d t=1, x>\frac{1}{2} $$

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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