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Use a table of integrals to determine the following indefinite integrals. $$\int \frac{d x}{x \sqrt{81-x^{2}}}$$

Short Answer

Expert verified
Question: Determine the indefinite integral of the following expression: $$\int \frac{d x}{x \sqrt{81-x^{2}}}$$ Answer: The indefinite integral of the given expression is: $$\int \frac{d x}{x \sqrt{81-x^{2}}} = \frac{1}{9}\sec^{-1}\left(\frac{9}{\sqrt{81-x^2}}\right) + C$$

Step by step solution

01

Identify the appropriate substitution

In this problem, we have a square root with the form $$\sqrt{a^2-x^2}$$ (Here, \(a^2 = 81\), so \(a = 9\)). According to the trigonometric substitution rules, we will substitute: $$x = a\sin(\theta) \Longrightarrow x = 9\sin(\theta)$$ Now, we also need to find the differential, \(dx\). To do this, we will differentiate the equation from above. $$\frac{dx}{d\theta} = 9\cos(\theta)$$ So we have, $$dx = 9\cos(\theta) d\theta$$
02

Substitute and simplify

Now we will substitute \(x = 9\sin(\theta)\) and \(dx = 9\cos(\theta)d\theta\) into the integral: $$\int \frac{1}{(9\sin(\theta))\sqrt{81-81\sin^2(\theta)}}9\cos(\theta) d\theta$$ Now simplify the integral: $$\int \frac{9\cos(\theta)}{9\sin(\theta)\sqrt{81(1-\sin^2(\theta))}} d\theta$$ Cancelling \(9\) and using the identity \(\cos^2(\theta)+\sin^2(\theta)=1\) we get: $$\int \frac{\cos(\theta)}{\sin(\theta)\sqrt{81\cos^2(\theta)}} d\theta$$ Simplify further: $$\int \frac{1}{\sin(\theta)\sqrt{81\cos^2(\theta)}} d\theta$$
03

Solve the integral using the table of integrals

Now, from the table of integrals we find the formula for the integral: $$\int \frac{1}{\sin(\theta)\sqrt{a^2\cos^2(\theta)}} d\theta = \frac{1}{a}\sec^{-1}\left(\frac{a}{x}\right) + C$$ Here, \(a = 9\) and \(x = 9\sin(\theta)\). Thus, our integral becomes: $$\frac{1}{9} \sec^{-1}\left(\frac{9}{9\sin(\theta)}\right) + C$$ Now, simplify: $$\frac{1}{9} \sec^{-1}\left(\frac{1}{\sin(\theta)}\right) + C$$
04

Replace the substitution

Now we need to replace the trigonometric substitution with the original variable \(x\). Recall that we used the substitution \(x = 9\sin(\theta)\). So we have: $$\sin(\theta) = \frac{x}{9}$$ Using the Pythagorean identity with a right triangle, we have: $$\cos(\theta) = \frac{\sqrt{81-x^2}}{9}$$ Now compute the secant function: $$\sec(\theta) = \frac{1}{\cos(\theta)} = \frac{9}{\sqrt{81-x^2}}$$ Finally, we can write the result as: $$\frac{1}{9} \sec^{-1}\left(\frac{9}{x}\right) + C = \frac{1}{9}\sec^{-1}\left(\frac{9}{\sqrt{81-x^2}}\right) + C$$ So, the indefinite integral is given by: $$\int \frac{d x}{x \sqrt{81-x^{2}}} = \frac{1}{9}\sec^{-1}\left(\frac{9}{\sqrt{81-x^2}}\right) + C$$

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Key Concepts

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

Trigonometric Substitution
The technique of trigonometric substitution is used to simplify integrals that contain radicals, as seen in the exercise \( \int \frac{dx}{x \sqrt{81-x^{2}}} \). By substituting a trigonometric function for the variable, the integral becomes easier to handle. For example, in cases that involve \( \sqrt{a^2-x^2} \) the substitution \( x = a \sin(\theta) \) is often used.

The strategy leverages the fundamental \( \sin^2(\theta) + \cos^2(\theta) = 1 \) Pythagorean identity, transforming the square root into a trigonometric expression that simplifies nicely. This method requires not only the substitution itself but also replacing the differential \( dx \) with \( a\cos(\theta)d\theta \) after differentiating \( x = a\sin(\theta) \) with respect to \( \theta \).
Integration Techniques
To solve complex integrals, various integration techniques may be employed. These include substitution, integration by parts, partial fraction decomposition, and tricky algebraic manipulations, among others. Each technique serves a purpose and excels at simplifying different forms of integrals.

The art of integration often lies in recognizing which technique to apply to a given problem. For instance, trigonometric substitution is perfect for dealing with square roots involving a difference of squares, while integration by parts is suitable for products of functions. Mastering these techniques is critical for success in calculus, especially when working with indefinite integrals that require anti-derivation.
Calculus
Calculus, the mathematical study of continuous change, is split into two main branches: differential calculus and integral calculus. Differential calculus concerns the calculation of derivatives—which measure rates of change—while integral calculus focuses on the concept of the integral, used to determine areas under curves, among other things.

Indefinite integrals are a part of integral calculus and represent a family of functions with an added constant \( C \), known as the constant of integration. Understanding the principles of calculus is essential to solve problems involving rates of change and accumulation, like the given exercise \( \int \frac{dx}{x \sqrt{81-x^{2}}} \).
Table of Integrals
A table of integrals is an invaluable tool in calculus, providing formulas for the integrals of various functions. These tables save time and effort by eliminating the need to solve common integral forms from scratch. Many of these tables have been developed and refined over centuries by mathematicians, and they often include a wide variety of trigonometric, exponential, logarithmic, and polynomial functions.

The table of integrals allows for pattern recognition—matching the integral at hand with a corresponding integral in the table to find its antiderivative. This method was applied in the solution where, after simplification through trigonometric substitution, the integral matched a standard form found in the table of integrals.
Pythagorean Identity
The Pythagorean identity, one of the most fundamental and frequently used identities in trigonometry, states that \( \sin^2(\theta) + \cos^2(\theta) = 1 \) for any angle \( \theta \). It's derived from the Pythagorean theorem applied to a right triangle with a hypotenuse of 1.

This identity is crucial in integrating functions involving square roots of expressions like \( \sqrt{a^2-x^2} \) because it allows for trigonometric substitutions that can simplify the integral. By substituting \( x = a\sin(\theta) \) or \( x = a\cos(\theta) \) in such expressions, one can convert the square root into a simpler trigonometric function, as demonstrated in the exercise's solution.

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Most popular questions from this chapter

An open cylindrical tank initially filled with water drains through a hole in the bottom of the tank according to Torricelli's Law (see figure). If \(h(t)\) is the depth of water in the tank for \(t \geq 0,\) then Torricelli's Law implies \(h^{\prime}(t)=2 k \sqrt{h}\), where \(k\) is a constant that includes the acceleration due to gravity, the radius of the tank, and the radius of the drain. Assume that the initial depth of the water is \(h(0)=H\). a. Find the general solution of the equation. b. Find the solution in the case that \(k=0.1\) and \(H=0.5 \mathrm{m}\). c. In general, how long does it take the tank to drain in terms of \(k\) and \(H ?\)

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Suppose \(f\) is positive and its first two derivatives are continuous on \([a, b] .\) If \(f^{\prime \prime}\) is positive on \([a, b],\) then is a Trapezoid Rule estimate of \(\int_{a}^{b} f(x) d x\) an underestimate or overestimate of the integral? Justify your answer using Theorem 7.2 and an illustration.

A differential equation of the form \(y^{\prime}(t)=F(y)\) is said to be autonomous (the function \(F\) depends only on \(y\) ). The constant function \(y=y_{0}\) is an equilibrium solution of the equation provided \(F\left(y_{0}\right)=0\) (because then \(y^{\prime}(t)=0,\) and the solution remains constant for all \(t\) ). Note that equilibrium solutions correspond to horizontal line segments in the direction field. Note also that for autonomous equations, the direction field is independent of \(t\). Consider the following equations. a. Find all equilibrium solutions. b. Sketch the direction field on either side of the equilibrium solutions for \(t \geq 0\). c. Sketch the solution curve that corresponds to the initial condition \(y(0)=1\). $$y^{\prime}(t)=y(y-3)$$

\(A n\) integrand with trigonometric functions in the numerator and denominator can often be converted to a rational integrand using the substitution \(u=\tan (x / 2)\) or equivalently \(x=2 \tan ^{-1} u .\) The following relations are used in making this change of variables. \(A: d x=\frac{2}{1+u^{2}} d u \quad B: \sin x=\frac{2 u}{1+u^{2}} \quad C: \cos x=\frac{1-u^{2}}{1+u^{2}}\) $$\text { Evaluate } \int_{0}^{\pi / 3} \frac{\sin \theta}{1-\sin \theta} d \theta$$

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