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Chapter 7: Problem 44

Evaluate the following integrals. $$\int \frac{d x}{x^{3} \sqrt{x^{2}-1}}, x>1$$

Short Answer

Expert verified
Question: Evaluate the integral \(\int \frac{dx}{x^3 \sqrt{x^2 - 1}}\). Answer: The integral evaluates to \(\arctan(\sqrt{x^2-1}) + C\), where C is the constant of integration.

Step by step solution

01

Identify the substitution method

To evaluate the integral, we will use a trigonometric substitution method as it can handle square roots effectively. We'll substitute \(x\) with a trigonometric function of another variable, say \(\theta\). Here, we notice the term \(\sqrt{x^2 - 1}\), which suggests using the substitution \(x = \cosh{\theta}\). This substitution provides a corresponding hyperbolic identity: \(\sinh^2{\theta} + 1 = \cosh^2{\theta}\).
02

Perform the substitution

Let's substitute \(x\) with \(\cosh{\theta}\). This will also require determining the differential \(dx\). We get: $$x = \cosh{\theta} \implies dx = \sinh{\theta} d\theta$$ Now, let's substitute these expressions into the integral: $$\int \frac{d x}{x^{3} \sqrt{x^{2}-1}}=\int \frac{\sinh{\theta} d\theta}{(\cosh^3{\theta}) \sqrt{\cosh^2{\theta}-1}}$$
03

Simplify the expression

Let's simplify the expression inside the integral using the identity \(\sinh^2{\theta} + 1=\cosh^2{\theta}\). This leads to the following: $$\int \frac{\sinh{\theta} d\theta}{(\cosh^3{\theta}) \sqrt{\sinh^2{\theta}}}$$ Now, we can simplify the expression further: $$\int \frac{\sinh{\theta} d\theta}{\cosh^3{\theta} \cdot \sinh{\theta}} = \int \frac{d\theta}{\cosh^3{\theta}}$$
04

Solve the integral

The integral now is easier to solve using the substitution \(y = \tanh{\theta}\), from which we get \(1-y^2 = \cosh^2{\theta}\) and \(dy = (\text{sech}^2{\theta})d\theta\). The integral becomes: $$\int \frac{d\theta}{(1-y^2)^{\frac{3}{2}}} = \int \frac{(\text{sech}^2{\theta})d\theta}{\text{sech}^3{\theta}} = \int \text{sech}{\theta} d\theta$$ Now, we know integration of \(\text{sech}{\theta}\) is \(\arctan(\sinh{\theta}) + C\). Thus, we have: $$\int \text{sech}{\theta} d\theta = \arctan(\sinh{\theta}) + C$$
05

Convert back to \(x\) variable

Now, let's convert back to the \(x\) variable. Recall that we have \(x = \cosh{\theta}\), and therefore, \(\sinh{\theta}=\sqrt{x^2-1}\). Finally, our solution is: $$\int \frac{d x}{x^{3} \sqrt{x^{2}-1}} = \arctan(\sqrt{x^2-1}) + C$$

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Key Concepts

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

Integrals
In calculus, an integral assigns numbers to functions in a way that can describe displacement, area, volume, and other concepts that arise from combining infinitesimal data. Integration is one of the two main operations in calculus, with its inverse operation being differentiation. The process of finding an integral is called integration. In technical language, an integral assigns a number to a function where that number often represents the area under the curve of the function on a graph.

To tackle more complex integrals, especially those involving square roots or special functions, we sometimes apply advanced techniques such as trigonometric substitution. This method is part of the integration by substitution strategy, where we replace the original variable with a trigonometric function of a new variable to simplify the integral. Trigonometric substitution is particularly useful when dealing with expressions under a square root, as it leverages the Pythagorean identity from trigonometry to convert the square root of a sum or difference of squares into a more workable form.
Hyperbolic Functions
The hyperbolic functions are analogs of the ordinary trigonometric, or circular, functions. Just as trigonometric functions are related to the circle, hyperbolic functions are related to the hyperbola. The two basic hyperbolic functions are the hyperbolic sine and cosine, denoted as \(\sinh{x}\) and \(\cosh{x}\), respectively. They are defined based on the exponential function as:
\[\sinh{x} = \frac{e^{x} - e^{-x}}{2}\] and \[\cosh{x} = \frac{e^{x} + e^{-x}}{2}\].

Important identities that relate these two functions include \(\cosh^2{x} - \sinh^2{x} = 1\) and the derivatives \(\frac{d}{dx} \sinh{x} = \cosh{x}\) and \(\frac{d}{dx} \cosh{x} = \sinh{x}\). These functions and their properties are particularly useful in solving certain types of integrals and differential equations that involve expressions reminiscent of the trigonometric Pythagorean identity.
Inverse Trigonometric Functions
The inverse trigonometric functions are the inverse functions of the trigonometric functions with restricted domains. In other words, they are used to find the angle with a given trigonometric ratio. There are six principal inverse trigonometric functions: \(\arcsin(x)\), \(\arccos(x)\), \(\arctan(x)\), \(\arccot(x)\), \(\arcsec(x)\), and \(\arccsc(x)\).

One particularly useful inverse trigonometric function in integration is the inverse tangent, \(\arctan(x)\), because it often appears as the result of an integral involving a quadratic expression in the denominator. This was indeed seen in the exercise solution, where the integration of hyperbolic secant \(\text{sech}(\theta)\) led to the inverse tangent \(\arctan(\sinh{\theta})\). Learning how to use and manipulate these functions is essential when dealing with the integrals that involve trigonometric expressions, especially as part of trigonometric substitutions.

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Most popular questions from this chapter

Use the indicated methods to solve the following problems with nonuniform grids. A hot-air balloon is launched from an elevation of 5400 ft above sea level. As it rises, its vertical velocity is computed using a device (called a variometer) that measures the change in atmospheric pressure. The vertical velocities at selected times are shown in the table (with units of \(\mathrm{ft} / \mathrm{min}\) ). $$\begin{array}{|l|c|c|c|c|c|c|c|} \hline t \text { (min) } & 0 & 1 & 1.5 & 3 & 3.5 & 4 & 5 \\ \hline \begin{array}{l} \text { Velocity } \\ \text { (ft/min) } \end{array} & 0 & 100 & 120 & 150 & 110 & 90 & 80 \\ \hline \end{array}$$ a. Use the Trapezoid Rule to estimate the elevation of the balloon after five minutes. Remember that the balloon starts at an elevation of \(5400 \mathrm{ft}\) b. Use a right Riemann sum to estimate the elevation of the balloon after five minutes. c. A polynomial that fits the data reasonably well is $$g(t)=3.49 t^{3}-43.21 t^{2}+142.43 t-1.75$$ Estimate the elevation of the balloon after five minutes using this polynomial.

\(A n\) 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. \(A: d x=\frac{2}{1+u^{2}} d u \quad B: \sin x=\frac{2 u}{1+u^{2}} \quad C: \cos x=\frac{1-u^{2}}{1+u^{2}}\) $$\text { Evaluate } \int_{0}^{\pi / 3} \frac{\sin \theta}{1-\sin \theta} d \theta$$

Evaluate the following integrals. Assume a and b are real numbers and \(n\) is an integer. $$\int \frac{x}{\sqrt{a x+b}} d x\left(\text { Use } u^{2}=a x+b\right.$$

Suppose that the rate at which a company extracts oil is given by \(r(t)=r_{0} e^{-k t},\) where \(r_{0}=10^{7}\) barrels \(/ \mathrm{yr}\) and \(k=0.005 \mathrm{yr}^{-1} .\) Suppose also the estimate of the total oil reserve is \(2 \times 10^{9}\) barrels. If the extraction continues indefinitely, will the reserve be exhausted?

Consider the general first-order initial value problem \(y^{\prime}(t)=a y+b, y(0)=y_{0},\) for \(t \geq 0,\) where \(a, b,\) and \(y_{0}\) are real numbers. a. Explain why \(y=-b / a\) is an equilibrium solution and corresponds to horizontal line segments in the direction field. b. Draw a representative direction field in the case that \(a>0\). Show that if \(y_{0}>-b / a,\) then the solution increases for \(t \geq 0\) and if \(y_{0}<-b / a,\) then the solution decreases for \(t \geq 0\). c. Draw a representative direction field in the case that \(a<0\). Show that if \(y_{0}>-b / a,\) then the solution decreases for \(t \geq 0\) and if \(y_{0}<-b / a,\) then the solution increases for \(t \geq 0\).
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