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Use a table of integrals to evaluate the following indefinite integrals. Some of the integrals require preliminary work, such as completing the square or changing variables, before they can be found in a table. $$\int \frac{d x}{x \sqrt{144-x^{2}}}$$

Short Answer

Expert verified
Now that we've solved the indefinite integral, we can create a short answer question based on the solution: Question: Find the integral of \(\int \frac{dx}{x \sqrt{144 - x^2}}\). Answer: \(-\frac{1}{12} \ln\left|\frac{12 + \sqrt{144 - x^2}}{x}\right| + C\)

Step by step solution

01

Perform substitution with a trigonometric function

To begin solving this integral, we will make a substitution using a trigonometric function to simplify the integrand. Since there's a square root involved, it's a good idea to use the sine function. We'll substitute as follows: $$x = 12 \sin{\theta}$$ Differentiating both sides with respect to \(x\): $$dx = 12 \cos{\theta} d\theta$$ Now replace \(x\) and \(dx\) in the integral with the substitutions.
02

Use the substitution and simplify

Substituting into the integral and simplifying: $$\int \frac{dx}{x \sqrt{144 - x^2}} = \int \frac{12\cos{\theta} d\theta}{12\sin{\theta} \sqrt{144 - (12\sin{\theta})^2}}$$ Simplify the expression: $$= \int \frac{\cos{\theta} d\theta}{\sin{\theta} \sqrt{144(1 - \sin^2{\theta})}}$$ Recall that \(1 - \sin^2{\theta} = \cos^2{\theta}\), so we have: $$= \int \frac{\cos{\theta} d\theta}{\sin{\theta} \sqrt{144\cos^2{\theta}}}$$
03

Calculate the integral

Cancelling the terms and taking constants out of the integral: $$= \frac{1}{12} \int \frac{\cos{\theta}}{\sin{\theta} \cdot \cos{\theta}} d\theta$$ $$= \frac{1}{12} \int \frac{d\theta}{\sin{\theta}}$$ Now use the table of integrals to find the integral of \(\frac{1}{\sin{\theta}}\): $$\int \frac{d\theta}{\sin{\theta}} = -\ln|\csc{\theta}+\cot{\theta}| + C$$
04

Reverse the substitution

Now reverse the substitution to find the solution in terms of \(x\). Recall that \(x = 12\sin{\theta}\). Using the Pythagorean identity, we have: $$\sin{\theta} = \frac{x}{12}$$ $$\cos{\theta} = \frac{\sqrt{144 - x^2}}{12}$$ Find \(\cot{\theta}\) and \(\csc{\theta}\): $$\cot{\theta} = \frac{\cos{\theta}}{\sin{\theta}} = \frac{\sqrt{144 - x^2}}{x}$$ $$\csc{\theta} = \frac{1}{\sin{\theta}} = \frac{12}{x}$$ Now substitute back into the integral: $$-\frac{1}{12} \ln|\frac{12}{x} + \frac{\sqrt{144 - x^2}}{x}| + C$$ Finally, the solution for the indefinite integral is: $$\int \frac{dx}{x \sqrt{144 - x^2}} = -\frac{1}{12} \ln\left|\frac{12 + \sqrt{144 - x^2}}{x}\right| + C$$

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Key Concepts

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

Trigonometric Substitution
Understanding trigonometric substitution is essential when dealing with integrals involving radicals. In cases such as \(\int \frac{dx}{x \sqrt{144-x^2}}\), where direct integration isn't immediately obvious, substituting a trigonometric function for a variable simplifies the problem.

The trigonometric substitution \(x = 12 \sin{\theta}\) strategically transforms the radical into a more manageable trigonometric identity. This method relies on Pythagorean identities like \(1 - \sin^2{\theta} = \cos^2{\theta}\), streamlining the integrand to involve basic trigonometric functions easy to find in standard integral tables.

As students approach this technique, it is helpful to visualize the right triangles and trigonometric relationships. For more complex expressions, drawing a triangle can often clarify which substitution to use and how to return to the original variable once the integral is evaluated. Since the goal is simplification, picking the right trig function—sine, cosine, or tangent—is crucial and depends on the form of the original integral.
Integrating Using Tables
Integrating using tables is a convenient approach to finding antiderivatives, especially when the integrand aligns well with known formulas. However, before looking up an integral, it is often necessary to do preliminary work such as completing the square or making a substitution (like trigonometric substitution, as mentioned earlier).

Once the integrand looks like a form found in a standard table of integrals, you can directly substitute the information into the provided formulas. For example, many tables include the integral \(\int \frac{d\theta}{\sin{\theta}}\), simplifying the integration process.

It’s important for students to familiarize themselves with these tables and understand how to navigate the entries. They provide a quick and effective method for solving complex integrals, but knowing when and how to use them is key. Also, flipping the substitutions 'back' illustrates the importance of keeping track of the initial variable, as the final answer must be expressed in terms of it.
Completing the Square
Completing the square is an algebraic technique that can be used to transform quadratic expressions into a perfect square plus a constant. This process often precedes integration as it simplifies the integrand into a form that may be more readily integrable.

While not explicitly shown in our example, if the integrand had been a more complex quadratic, we could have resorted to completing the square to simplify the expression. The modified integrand can then either be integrated directly or further processed with a variable or trigonometric substitution.

For students, mastering completing the square offers a way to tackle various quadratic integrands and is also beneficial for understanding the derivation of the quadratic formula and for solving certain differential equations. The aim here is to recognize when completing the square can turn an otherwise daunting integral into something much more approachable.
Variable Change in Integration
A variable change in integration, often known as u-substitution, is a powerful tool for simplifying integrals. It involves substituting a part of the integrand with a new variable 'u' and its differential 'du' to make the integral easier to evaluate.

In the given problem, we made a trigonometric substitution to deal with the square root, where \(x\) was replaced with \(12 \sin{\theta}\) and \(dx\) with \(12 \cos{\theta} d\theta\). This substitution itself is an example of changing variables to simplify the integrand.

Students should note that the choice of substitution is not always straightforward and may require several attempts or different strategies for different integrals. The substitution made should simplify the integral into a form that is either straightforward to integrate or matches an entry in a table of standard integrals. In essence, the art of variable change is like finding the key to unlock a complex integral, making it more accessible and solvable.

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Most popular questions from this chapter

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. a. It is possible for a computer algebra system to give the result \(\int \frac{d x}{x(x-1)}=\ln (x-1)-\ln x\) and a table of integrals to give the result \(\int \frac{d x}{x(x-1)}=\ln \left|\frac{x-1}{x}\right|+C\) b. A computer algebra system working in symbolic mode could give the result \(\int_{0}^{1} x^{8} d x=\frac{1}{9},\) and a computer algebra system working in approximate (numerical) mode could give the result \(\int_{0}^{1} x^{8} d x=0.11111111\).

Practice with tabular integration Evaluate the following integrals using tabular integration (refer to Exercise 77 ). a. \(\int x^{4} e^{x} d x \quad\) b. \(\int 7 x e^{3 x} d x\) c. \(\int_{-1}^{0} 2 x^{2} \sqrt{x+1} d x\) d. \(\int\left(x^{3}-2 x\right) \sin 2 x \, d x\) e. \(\int \frac{2 x^{2}-3 x}{(x-1)^{3}} d x\) f. \(\int \frac{x^{2}+3 x+4}{\sqrt[3]{2 x+1}} d x\) g. Why doesn't tabular integration work well when applied to \(\int \frac{x}{\sqrt{1-x^{2}}} d x \, ?\) Evaluate this integral using a different method.

The nucleus of an atom is positively charged because it consists of positively charged protons and uncharged neutrons. To bring a free proton toward a nucleus, a repulsive force \(F(r)=k q Q / r^{2}\) must be overcome, where \(q=1.6 \times 10^{-19} \mathrm{C}(\) coulombs ) is the charge on the proton, \(k=9 \times 10^{9} \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}, Q\) is the charge on the nucleus, and \(r\) is the distance between the center of the nucleus and the proton. Find the work required to bring a free proton (assumed to be a point mass) from a large distance \((r \rightarrow \infty)\) to the edge of a nucleus that has a charge \(Q=50 q\) and a radius of \(6 \times 10^{-11} \mathrm{m}\).

Three cars, \(A, B,\) and \(C,\) start from rest and accelerate along a line according to the following velocity functions: $$v_{A}(t)=\frac{88 t}{t+1}, \quad v_{B}(t)=\frac{88 t^{2}}{(t+1)^{2}}, \quad \text { and } \quad v_{C}(t)=\frac{88 t^{2}}{t^{2}+1}$$ a. Which car travels farthest on the interval \(0 \leq t \leq 1 ?\) b. Which car travels farthest on the interval \(0 \leq t \leq 5 ?\) c. Find the position functions for each car assuming each car starts at the origin. d. Which car ultimately gains the lead and remains in front?

Show that \(\int_{0}^{\infty} \frac{\sqrt{x} \ln x}{(1+x)^{2}} d x=\pi\) in the following steps. a. Integrate by parts with \(u=\sqrt{x} \ln x\). b. Change variables by letting \(y=1 / x\). c. Show that \(\int_{0}^{1} \frac{\ln x}{\sqrt{x}(1+x)} d x=-\int_{1}^{\infty} \frac{\ln x}{\sqrt{x}(1+x)} d x\) (and that both integrals converge). Conclude that \(\int_{0}^{\infty} \frac{\ln x}{\sqrt{x}(1+x)} d x=0\). d. Evaluate the remaining integral using the change of variables \(z=\sqrt{x}\). (Source: Mathematics Magazine \(59,1,\) Feb 1986 )

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