/*! 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 13 Evaluate the following integrals... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 13

Evaluate the following integrals: $$ \int \frac{x^{2} d x}{\sqrt{x^{2}-16}} $$

Short Answer

Expert verified
Question: Evaluate the integral: $$\int\frac{x^2}{\sqrt{x^2-16}}dx$$ Answer: $$4\sqrt{x^2-16}+C$$

Step by step solution

01

Perform Trigonometric Substitution

Let's start the trigonometric substitution by letting: $$ x = 4\sec(\theta) $$ From this substitution, the derivative of x with respect to 胃 is: $$ dx = 4\sec(\theta)\tan(\theta)d\theta $$ Now, let's adjust the expression inside the square root: $$ \sqrt{x^2 - 16} = \sqrt{(4\sec(\theta))^2 - 16} $$ Simplify the expression: $$ \sqrt{16\sec^2(\theta) - 16} = 4\sqrt{\sec^2(\theta) - 1} $$ Since \(\sec^2(\theta) - 1 = \tan^2(\theta)\), then: $$ \sqrt{x^2 - 16} = 4\tan(\theta). $$
02

Rewrite Integral in Terms of Substituted Variable

Now let's replace the expression with the substitution: $$ \int \frac{x^2}{\sqrt{x^2-16}}dx = \int \frac{(4\sec(\theta))^2}{4\tan(\theta)}(4\sec(\theta)\tan(\theta)d\theta) $$ Simplify the integral: $$ \int 16\sec^2(\theta)d\theta $$
03

Compute Integral

Compute the integral of the simplified expression: $$ \int 16\sec^2(\theta)d\theta = 16\int\sec^2(\theta)d\theta $$ The integral of \(\sec^2(\theta)\) is \(\tan(\theta)\), so: $$ 16\int\sec^2(\theta)d\theta = 16\tan(\theta) + C $$ Where C is the constant of integration.
04

Transform back to Original Variable

Now, we'll replace 胃 with the original variable x. Recall our original substitution: $$ x = 4\sec(\theta) $$ Now, we'll use the fact that \(\sec^2(\theta) - 1 = \tan^2(\theta)\): $$ \tan(\theta) = \sqrt{\sec^2(\theta) - 1} = \frac{\sqrt{x^2 -16}}{4} $$ Thus, the final answer is: $$ 16\tan(\theta) + C = 16\left(\frac{\sqrt{x^2 -16}}{4}\right) + C = 4\sqrt{x^2 -16} + C $$

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

Evaluate the following integrals: (i) \(\int \frac{2 x^{3}+x^{2}+4}{\left(x^{2}+4\right)^{2}} d x\) (ii) \(\int \frac{x^{3}+x^{2}-5 x+15}{\left(x^{2}+5\right)\left(x^{2}+2 x+3\right)} d x\)(iii) \(\int \frac{d x}{\left(x^{4}+2 x+10\right)^{3}}\) (iv) \(\int \frac{x^{5}-x^{4}+4 x^{3}-4 x^{2}+8 x-4}{\left(x^{2}+2\right)^{3}} d x\)

If \(I_{n}=\int \frac{x^{n}}{\sqrt{x^{2}+a^{2}}} d x(n \geq 2)\), then show that \(I_{n}=\frac{x^{n-1} \sqrt{x^{2}+a^{2}}}{n}-\frac{a^{2}(n-1)}{n} I_{n-2}\)

Deduce the reduction formula for \(I_{n}=\int \frac{d x}{\left(1+x^{4}\right)^{n}}\) andhenceevaluate \(I_{2}=\int \frac{d x}{\left(1+x^{4}\right)^{2}} .\)

Evaluate the following integrals: $$ \int \frac{\sqrt{2+x^{2}}-\sqrt{2-x^{2}}}{\sqrt{4-x^{4}}} d x $$

\(\int \frac{x^{2}-7 x+1}{\sqrt[3]{2 x+1}} d x\)
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Mechanics Maths

Read Explanation

Decision Maths

Read Explanation

Pure Maths

Read Explanation

Theoretical and Mathematical Physics

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.