/*! 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 27 Evaluate the following integrals... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 27

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

Short Answer

Expert verified
Question: Evaluate the following integral: $$\int \frac{2 - 3x}{\sqrt{1 - x^2}} dx$$ Answer: $$\int \frac{2 - 3x}{\sqrt{1 - x^2}} dx = 2\arcsin{x} + 3\sqrt{1 - x^2} + C$$

Step by step solution

01

Separate the fraction into two simpler fractions

We can rewrite the integrand as follows: $$ \frac{2 - 3x}{\sqrt{1 - x^2}} = \frac{2}{\sqrt{1 - x^2}} - \frac{3x}{\sqrt{1 - x^2}} $$ Now, we have two simpler fractions to integrate separately.
02

Use trigonometric substitution

In order to solve these integrals, we will use trigonometric substitution. Let's set \(x = \sin{u}\), so that \(dx = \cos{u}\ du\). We also have \(\sqrt{1 - x^2} = \cos{u}\). Now, let's substitute these values into our integrals: $$\int \frac{2 - 3x}{\sqrt{1 - x^2}} dx = \int \frac{2}{\cos{u}} \cos{u}\ du - \int \frac{3 \sin{u}}{\cos{u}} \cos{u}\ du$$
03

Simplify and integrate

Now we can simplify the integrals and integrate each term with respect to \(u\): $$\int \frac{2}{\cos{u}} \cos{u}\ du - \int \frac{3 \sin{u}}{\cos{u}} \cos{u}\ du = \int 2 \ du - \int 3 \sin{u} \ du$$ $$ = 2u - (-3\cos{u}) + C = 2u + 3\cos{u} + C$$
04

Convert back to \(x\)

We are now done with the integration, but we must express the result in terms of \(x\) rather than \(u\). Remember that we made the substitution \(x = \sin{u}\), so \(u = \arcsin{x}\): $$2u + 3\cos{u} + C = 2\arcsin{x} + 3\cos{\arcsin{x}} + C$$ To find \(\cos{\arcsin{x}}\), we can use the Pythagorean identity \(\sin^2{u} + \cos^2{u} = 1\): $$\cos^2{u} = 1 - \sin^2{u} = 1 - x^2$$ $$\cos{u} = \sqrt{1 - x^2}$$ Now, we can substitute this back into our expression: $$2\arcsin{x} + 3\cos{\arcsin{x}} + C = 2\arcsin{x} + 3\sqrt{1 - x^2} + C$$ So, the final result is: $$\int \frac{2 - 3x}{\sqrt{1 - x^2}} dx = 2\arcsin{x} + 3\sqrt{1 - x^2} + C$$

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!

Key Concepts

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

trigonometric substitution
Trigonometric substitution is a powerful technique used in calculus to tackle integrals involving square roots of quadratic expressions. This method involves substituting a trigonometric function for an algebraic variable to simplify the integral. In our exercise, we used the substitution \( x = \sin{u} \), which is especially handy for integrals involving \( \sqrt{1-x^2} \). This substitution works well because \( \cos^2{u} + \sin^2{u} = 1 \) is the Pythagorean identity that simplifies the expression under the square root.
  • The relation \( \sqrt{1-x^2} = \cos{u} \) helps us handle the square root easily.
  • Substituting \( x = \sin{u} \) implies \( dx = \cos{u} \ du \), which is crucial for transforming the integral.
Trigonometric substitution transforms the variables and consequently simplifies the integration process.
integration by parts
Integration by parts is another essential technique in calculus, especially useful for integrals that are products of two functions. The formula for integration by parts is derived from the product rule for differentiation and is given by: \[ \int u \ dv = uv - \int v \ du \] This method is particularly helpful for integrating functions that are difficult to integrate directly but are simpler when broken down.
  • The exercise did not require integration by parts explicitly, but understanding it complements our approach.
  • You would use it when facing integrals of products that aren't straightforward, such as \( \int xe^x dx \).
Breaking down the integral into manageable parts is a strategic approach in problem-solving.
Pythagorean identity
The Pythagorean identity is a fundamental relation in trigonometry. It states that for any angle \( u \), the equation \( \sin^2{u} + \cos^2{u} = 1 \) holds true. This identity is extremely useful in simplifying expressions involving trigonometric functions.
  • We used this identity in our exercise to replace \( \sqrt{1-x^2} \) with \( \cos{u} \).
  • This simplified the integral because it connected trigonometric substitution directly with the properties of sine and cosine.
Knowing Pythagorean identities equips you with the ability to simplify complex trigonometric expressions, making problems more approachable.

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. $$\int \frac{d x}{1-\tan ^{2} x}$$

sine integral The theory of diffraction produces the sine integral function \(\mathrm{Si}(x)=\int_{0}^{x} \frac{\sin t}{t} d t .\) Use the Midpoint Rule to approximate \(\left.\operatorname{Si}(1) \text { and } \operatorname{Si}(10) . \text { (Recall that } \lim _{x \rightarrow 0} \frac{\sin x}{x}=1 .\right)\) Experiment with the number of subintervals until you obtain approximations that have an error less than \(10^{-3}\). A rule of thumb is that if two successive approximations differ by less than \(10^{-3}\), then the error is usually less than \(10^{-3} .\)

Let \(a>0\) and \(b\) be real numbers. Use integration to confirm the following identities. (See Exercise 73 of Section 8.2) a. \(\int_{0}^{\infty} e^{-a x} \cos b x d x=\frac{a}{a^{2}+b^{2}}\) b. \(\int_{0}^{\infty} e^{-a x} \sin b x d x=\frac{b}{a^{2}+b^{2}}\)

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 s is a positive real number. For example, to find the Laplace transform of \(f(t)=e^{-t},\) the following improper integral is evaluated using integration by parts: $$F(s)=\int_{0}^{\infty} e^{-s t} e^{-t} d t=\int_{0}^{\infty} e^{-(s+1) t} d t=\frac{1}{s+1}$$ Verify the following Laplace transforms, where a is a real number. $$f(t)=\sin a t \rightarrow F(s)=\frac{a}{s^{2}+a^{2}}$$

Evaluate the following integrals. $$\int \frac{\tan \theta+\tan ^{3} \theta}{(1+\tan \theta)^{50}} d \theta$$
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Logic and Functions

Read Explanation

Applied Mathematics

Read Explanation

Statistics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Mechanics Maths

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.