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Evaluate the following integrals. $$\int \frac{d x}{\sqrt{x^{2}-49}}, x>7$$

Short Answer

Expert verified
Question: Evaluate the integral $$\int \frac{dx}{\sqrt{x^2 - 49}}.$$ Answer: The integral evaluates to $$\text{arcosh}\left(\frac{x}{7}\right) + C,$$ where \(C\) is the constant of integration.

Step by step solution

01

Identify the appropriate substitution

Since this integral resembles the inverse hyperbolic function, let's use the substitution: $$x = 7\cosh{u}.$$ Then, the differential will be: $$dx = 7\sinh{u} du.$$
02

Substitute and Simplify

Now, substitute the expressions for \(x\) and \(dx\) into the integral: $$\int \frac{d x}{\sqrt{x^{2}-49}} = \int \frac{7\sinh{u} du}{\sqrt{(7\cosh{u})^2 - 49}}.$$ By plugging the substitution, we need to simplify the expression inside the square root: $$\sqrt{(7\cosh{u})^2 - 49} = \sqrt{49\cosh^2{u} - 49} = \sqrt{49(\cosh^2{u}-1)}.$$ Now we know that \(\cosh^2{u}-1 = \sinh^2{u}\). So, the expression inside the square root becomes: $$\sqrt{49(\cosh^2{u}-1)} = \sqrt{49\sinh^2{u}} = 7\sinh{u}.$$
03

Integrate

Substitute the simplified expression back into the integral and integrate: $$\int \frac{7\sinh{u} du}{7\sinh{u}} = \int du.$$ Integrating w.r.t \(u\) gives: $$u+C,$$ where \(C\) is the constant of integration.
04

Substitute back

Now, substitute back the original variable \(x\) into the expression: $$u+C = \text{arcosh}\left(\frac{x}{7}\right)+C.$$ So, the final result is: $$\int \frac{d x}{\sqrt{x^{2}-49}} = \text{arcosh}\left(\frac{x}{7}\right)+C.$$

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Key Concepts

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

Inverse Hyperbolic Functions
Inverse hyperbolic functions, such as \text{arcosh}, \text{arsinh}, and \text{artanh}, are the inverses of the standard hyperbolic functions: \text{cosh}, \text{sinh}, and \text{tanh}. These functions can be quite useful when solving integrals involving radical expressions, especially those that resemble the forms of hyperbolic functions.

\text{arcosh} (area hyperbolic cosine) is the inverse function of \text{cosh} and is defined only for values greater than or equal to 1. In the context of integration, it often appears when we encounter an integral of the form \( \int \frac{dx}{\sqrt{x^2 - a^2}} \) where \( x > a \).

For example, to evaluate the indefinite integral given in the exercise, the substitution \( x = a\cosh{u} \) is employed because it simplifies the integration process. The differential \( dx \) is then expressed in terms of \( du \) using the derivative of \( \cosh{u} \) which is \( \sinh{u} \) and the associative constants.

It’s worth noting that the use of these functions greatly simplifies the integral, transforming a potentially complex radical expression into a more manageable form. This approach is often much faster and cleaner than alternative methods such as trigonometric substitution.
Substitution Method in Integration
The substitution method, also known as u-substitution, is a powerful technique for solving integrals. This approach involves replacing a segment of the integral with a new variable \( u \), which simplifies the expression into an easier form to integrate. It is analogous to the chain rule in differentiation.

Steps of Substitution:

  • Choose a substitution that simplifies the integral.
  • Express the differential \( dx \) in terms of \( du \).
  • Rewrite the integral with the new variable \( u \) and find the antiderivative.
  • Convert back to the original variable.
In the given exercise, the substitution method is effectively used by letting \( x \)=\( 7\cosh{u} \) where \( u \) is a new integration variable. By doing so, the integral simplifies to a form where we can easily find the antiderivative. Finally, the inverse substitution is performed to return to the original variable \( x \) yielding the solution in terms of inverse hyperbolic functions. The substitution method not only helps in simplification but also allows us to recognize and utilize certain function properties for integration.
Hyperbolic Trigonometric Identities
Hyperbolic trigonometric identities are relations between hyperbolic functions that allow for the simplification of complex expressions. These identities are similar in form to their trigonometric counterparts but differ in the signs of their terms.

One of the most useful hyperbolic identities is \( \cosh^2{u} - \sinh^2{u} = 1 \), which arises from the definition of hyperbolic functions based on exponential functions. An immediate consequence of this identity, which we use in the exercise, is \( \cosh^2{u} - 1 = \sinh^2{u} \).

Utilizing these identities not only simplifies the integral calculation but also ensures that the integrand is expressed in its simplest form. In our example, after performing the substitution \( x = 7\cosh{u} \), we employ the identity to simplify \( \sqrt{(7\cosh{u})^2 - 49} \) to \( 7\sinh{u} \) under the square root. This simplification is pivotal in moving forward with the integration process. Knowledge of these identities is essential for efficiently working with hyperbolic functions while integrating.

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Most popular questions from this chapter

Apply Simpson's Rule to the following integrals. It is easiest to obtain the Simpson's Rule approximations from the Trapezoid Rule approximations, as in Example \(7 .\) Make \(a\) table similar to Table 7.8 showing the approximations and errors for \(n=4,8,16,\) and \(32 .\) The exact values of the integrals are given for computing the error. \(\int_{0}^{4}\left(3 x^{5}-8 x^{3}\right) d x=1536\)

Let \(a>0\) and let \(R\) be the region bounded by the graph of \(y=e^{-a x}\) and the \(x\) -axis on the interval \([b, \infty)\) a. Find \(A(a, b),\) the area of \(R\) as a function of \(a\) and \(b\) b. Find the relationship \(b=g(a)\) such that \(A(a, b)=2\) c. What is the minimum value of \(b\) (call it \(b^{*}\) ) such that when \(b>b^{*}, A(a, b)=2\) for some value of \(a>0 ?\)

Use integration by parts to evaluate the following integrals. $$\int_{1}^{\infty} \frac{\ln x}{x^{2}} d x$$

The Eiffel Tower property Let \(R\) be the region between the curves \(y=e^{-\alpha x}\) and \(y=-e^{-\alpha x}\) on the interval \([a, \infty),\) where \(a \geq 0\) and \(c>0 .\) The center of mass of \(R\) is located at \((\bar{x}, 0)\) where \(\bar{x}=\frac{\int_{a}^{\infty} x e^{-c x} d x}{\int_{a}^{\infty} e^{-c x} d x} .\) (The profile of the Eiffel Tower is modeled by the two exponential curves; see the Guided Project The exponential Eiffel Tower.) a. For \(a=0\) and \(c=2,\) sketch the curves that define \(R\) and find the center of mass of \(R\). Indicate the location of the center of mass. b. With \(a=0\) and \(c=2,\) find equations of the lines tangent to the curves at the points corresponding to \(x=0\) c. Show that the tangent lines intersect at the center of mass. d. Show that this same property holds for any \(a \geq 0\) and any \(c>0 ;\) that is, the tangent lines to the curves \(y=\pm e^{-c x}\) at \(x=a\) intersect at the center of mass of \(R\).

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 \(y(0)=1\). $$y^{\prime}(t)=y^{2}$$

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