/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 15 Use a table of integrals to dete... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 15

Use a table of integrals to determine the following indefinite integrals. $$\int \frac{d x}{\sqrt{9 x^{2}-100}}, x>\frac{10}{3}$$

Short Answer

Expert verified
Question: Calculate the indefinite integral $$\int \frac{dx}{\sqrt{9x^2 - 100}}$$ Answer: The indefinite integral is $$\int \frac{1}{\sqrt{9x^2 - 100}} dx = \frac{10}{9} \cosh^{-1}{\left(\frac{9x}{100}\right)} + C$$

Step by step solution

01

Rewrite the integral with constants a and b

First, we rewrite the given integral, recognizing the a and b constants: $$\int \frac{dx}{\sqrt{9x^2 - 100}} = \int \frac{1}{\sqrt{9x^2 - 100}} dx$$
02

Identify the appropriate trigonometric substitution

For the integral of the form \(\int \frac{1}{\sqrt{ax^2 - b^2}} dx\), the appropriate substitution is \(x = \frac{b}{a}\cosh{u}\) with \(a=9\) and \(b=100\). The derivative of this substitution is \(dx = \frac{b}{a}\sinh{u} du\), where \(u\) is a new variable.
03

Apply the substitution

Now, apply the substitution to the integral: $$ \int \frac{1}{\sqrt{9x^2 - 100}} dx = \int \frac{1}{\sqrt{(9\cdot \frac{100}{9}\cosh{u})^2 - 100}} \cdot \frac{100}{9}\sinh{u} du$$
04

Simplify the integral

Simplify the expression within the square root and outside the integral: $$\int \frac{1}{\sqrt{9x^2 - 100}} dx = \int \frac{\frac{100}{9}\sinh{u}}{\sqrt{100(\cosh^2{u} - 1)}} du$$ Since \(100(\cosh^2{u} - 1) = 100\sinh^2{u}\), we can rewrite the above expression as: $$\int \frac{1}{\sqrt{9x^2 - 100}} dx = \int \frac{\frac{100}{9}\sinh{u}}{\sqrt{100\sinh^2{u}}} du$$
05

Simplify the integral further

Simplify again, we get: $$\int \frac{1}{\sqrt{9x^2 - 100}} dx = \int \frac{10}{9} du$$
06

Evaluate the integral

Now we have a straightforward integral: $$\int \frac{1}{\sqrt{9x^2 - 100}} dx = \frac{10}{9} \int dx = \frac{10}{9} u + C$$
07

Convert back to the original variable

Finally, we convert back to the original variable \(x\) using the substitution \(x = \frac{100}{9}\cosh{u}\): $$u = \cosh^{-1}{\left(\frac{9x}{100}\right)}$$ So, the solution to the problem is: $$\int \frac{1}{\sqrt{9x^2 - 100}} dx = \frac{10}{9} \cosh^{-1}{\left(\frac{9x}{100}\right)} + C$$

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

\(A\) powerful tool in solving problems in engineering and physics is the Laplace transform. Given a function \(f(t),\) the Laplace transform is a new function \(F(s)\) defined by $$ F(s)=\int_{0}^{\infty} e^{-s t} f(t) d t $$ where we assume that s is a positive real number. For example, to find the Laplace transform of \(f(t)=e^{-t},\) the following improper integral is evaluated: $$ F(s)=\int_{0}^{\infty} e^{-s t} e^{-t} d t=\int_{0}^{\infty} e^{-(s+1) t} d t=\frac{1}{s+1} $$ Verify the following Laplace transforms, where a is a real number. $$f(t)=e^{a t} \longrightarrow F(s)=\frac{1}{s-a}$$

An 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 \(x=2 \tan ^{-1} u .\) The following relations are used in making this change of variables. $$A: d x=\frac{2}{1+u^{2}} d u \quad B: \sin x=\frac{2 u}{1+u^{2}} \quad C: \cos x=\frac{1-u^{2}}{1+u^{2}}$$ $$\text { Evaluate } \int \frac{d x}{1+\sin x}$$

Graph the function \(f(x)=\frac{\sqrt{x^{2}-9}}{x}\) and consider the region bounded by the curve and the \(x\) -axis on \([-6,-3] .\) Then evaluate \(\int_{-6}^{-3} \frac{\sqrt{x^{2}-9}}{x} d x .\) Be sure the result is consistent with the graph.

The following integrals require a preliminary step such as long division or a change of variables before using partial fractions. Evaluate these integrals. $$\int \frac{e^{x}}{\left(e^{x}-1\right)\left(e^{x}+2\right)} d x$$

The following integrals require a preliminary step such as long division or a change of variables before using partial fractions. Evaluate these integrals. $$\int \frac{3 x^{2}+4 x-6}{x^{2}-3 x+2} d x$$
See all solutions

Recommended explanations on Math Textbooks

Mechanics Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Pure Maths

Read Explanation

Applied Mathematics

Read Explanation

Decision Maths

Read Explanation

Calculus

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.