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Chapter 7: Problem 38

Evaluate the following integrals. $$\int \frac{\sqrt{9-x^{2}}}{x^{2}} d x$$

Short Answer

Expert verified
Short answer: The definite integral: $$\int \frac{\sqrt{9-x^{2}}}{x^{2}} d x$$ can be simplified using the trigonometric substitution: $$x = 3 \sin(\theta)$$ After substitution, simplification, and reverse substitution, the integral becomes: $$-3 \int \frac{\left(\sqrt{9-x^2}\right)^3}{x^2}d(\arcsin(\frac{x}{3}))$$ The original integral is equal to the negative of this final integral, which can be evaluated using a table or WolframAlpha.

Step by step solution

01

Choose the substitution

Let's use the following trigonometric substitution: $$x = 3 \sin(\theta)$$ Now, we need to find \(\frac{dx}{d\theta}\): $$\frac{dx}{d\theta} = 3 \cos(\theta)$$
02

Rewrite the integral with the substitution

We rewrite the original integral by substituting the chosen substitutions: $$\int \frac{\sqrt{9-x^2}}{x^2}dx$$ $$\int \frac{\sqrt{9-(3\sin\theta)^2}}{(3\sin\theta)^2}(3\cos\theta)d\theta$$
03

Simplify the integral

We can now simplify the integral by factoring out constants, and utilizing trigonometric identities: $$\int \frac{\sqrt{9-(9\sin^2\theta)}}{9\sin^2\theta}(3\cos\theta)d\theta$$ $$\frac{1}{3}\int \frac{\sqrt{9(1-\sin^2\theta)}}{9\sin^2(\theta)}\cdot 3\cos(\theta)d\theta$$ According to the identity \(\cos^2(\theta) = 1 - \sin^2(\theta)\) $$\frac{1}{3}\int \frac{\sqrt{9\cos^2(\theta)}}{9\sin^2(\theta)}\cdot 3\cos(\theta)d\theta$$ $$\int \frac{3\cos^2(\theta)}{\sin^2(\theta)}\cos(\theta)d\theta$$ $$3\int \frac{\cos^3(\theta)}{\sin^2(\theta)}d\theta$$
04

Substitute back to x

Right now, we could evaluate the integral with the help of a table. However, in this case, we can perform the reverse substitution to get back to x. To do this, we need to remember that: $$x = 3\sin(\theta)$$ $$\sin(\theta) = \frac{x}{3}$$ We also need to find \(\cos^3\theta\): $$\cos(\theta) = \sqrt{1-\sin^2(\theta)}$$ $$\cos(\theta) = \sqrt{1-\left(\frac{x}{3}\right)^2}$$ $$\cos^3(\theta) = \left(\sqrt{9-x^2}\right)^3$$ And finally, we need the reverse substitution for \(\sin^2\theta\): $$\sin^2(\theta) = \left(\frac{x}{3}\right)^2$$ Substitute these back into the integral: $$3\int \frac{\left(\sqrt{9-x^2}\right)^3}{\left(\frac{x}{3}\right)^2}d(-\arcsin(\frac{x}{3}))$$
05

Cancel out the factor

Now, simplify the integral: $$-3\int \frac{\left(\sqrt{9-x^2}\right)^3}{x^2}d(\arcsin(\frac{x}{3}))$$ The integral is equal to the original integral multiplied by \(-1\). Therefore, the original integral is equal to the negative of this integral. This final integral corresponds to the one looked up in tables. Or you can use WolframAlpha, which is a nice tool for practicing and checking your solution.

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Key Concepts

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

Calculus
Calculus, the mathematical study of continuous change, underpins many of the laws of physics and is foundational to the science of mechanics. In the context of our integral problem, calculus provides us with tools like differentiation and integration to solve complex problems involving rates of change and areas under curves. An integral represents the accumulation of quantities, such as areas under a curve, and is used for finding values that are not easily computed through basic operations. In this exercise, the goal is to evaluate a definite integral with a square root and a trigonometric component, which are common in calculus problems involving shapes or motion, using the methods of trigonometric substitution and integral evaluation.
Trigonometric Identities
Trigonometric identities are powerful tools used to simplify and solve equations involving trigonometric functions. One such identity involves the Pythagorean identity, \(\cos^2(\theta) = 1 - \sin^2(\theta)\). This identity helps us transform an integral in a way that makes it more manageable to evaluate. In our exercise, we use this identity to rewrite the integral in terms of \(\cos\theta\) only, reducing complexity and enabling further simplification. The ability to recognize when and how to apply these identities is a key skill in successfully solving integrals involving trigonometric functions.
Integration by Substitution
Integration by substitution, also known as u-substitution, is a technique to evaluate integrals that are not immediately straightforward. Similar to the substitution method used for solving algebraic equations, this method involves changing the variable of integration to simplify the integral. In trigonometric substitution, as seen in this exercise, we replace the variable with a trigonometric expression that simplifies the radical and denominator. The process includes finding the differential of this substitution \(dx = 3\cos(\theta)d\theta\), which allows us to rewrite the entire integral in a more solvable form. Substitution is especially useful when dealing with integrals that contain roots or other complex expressions.
Integral Evaluation
Integral evaluation is the final step of the problem-solving process in calculus. It involves finding the antiderivative or calculating the definite integral to obtain the final answer. After simplifying the integral using trigonometric identities and substitution, we need to evaluate the resulting expression. It sometimes requires additional strategies such as integration by parts, partial fractions, or looking up integral tables. In our example, after simplification and substitution, the result is an integral that we can evaluate to find the area under the curve represented by the original function. The process highlights the importance of reversing the substitution to express the final answer in terms of the original variable.

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Most popular questions from this chapter

An endowment is an investment account in which the balance ideally remains constant and withdrawals are made on the interest earned by the account. Such an account may be modeled by the initial value problem \(B^{\prime}(t)=a B-m\) for \(t \geq 0,\) with \(B(0)=B_{0} .\) The constant \(a\) reflects the annual interest rate, \(m\) is the annual rate of withdrawal, and \(B_{0}\) is the initial balance in the account. a. Solve the initial value problem with a=0.05, m= 1000 dollar \(/\mathrm{yr}\), and \(B_{0}\)= 15,000 dollar. Does the balance in the account increase or decrease? b. If \(a=0.05\) and \(B_{0}\)= 50,000 dollar, what is the annual withdrawal rate \(m\) that ensures a constant balance in the account? What is the constant balance?

Given a Midpoint Rule approximation \(M(n)\) and a Trapezoid Rule approximation \(T(n)\) for a continuous function on \([a, b]\) with \(n\) subintervals, show that \(T(2 n)=(T(n)+M(n)) / 2\).

Let \(R\) be the region bounded by the graph of \(f(x)=x^{-p}\) and the \(x\) -axis, for \(x \geq 1\) a. Let \(S\) be the solid generated when \(R\) is revolved about the \(x\) -axis. For what values of \(p\) is the volume of \(S\) finite? b. Let \(S\) be the solid generated when \(R\) is revolved about the \(y\) -axis. For what values of \(p\) is the volume of \(S\) finite?

Graph the integrands and then evaluate and compare the values of \(\int_{0}^{\infty} x e^{-x^{2}} d x\) and \(\int_{0}^{\infty} x^{2} e^{-x^{2}} d x\).

Solve the following problems using the method of your choice. $$\frac{d p}{d t}=\frac{p+1}{t^{2}}, p(1)=3$$
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