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Chapter 8: Problem 46

Use an appropriate substitution and then a trigonometric substitution to evaluate the integrals. $$\begin{array}{l}{\int \sqrt{\frac{x}{1-x^{3}}} d x} \\ {\left(\text {Hint: Let } u=x^{3 / 2} .\right)}\end{array}$$

Short Answer

Expert verified
The integral evaluates to \( \arcsin(x^{3/2}) + C \).

Step by step solution

01

Substitution Introduction

First, we'll use the hint and substitute \( u = x^{3/2} \). This implies \( x = u^{2/3} \). We also need to change the differential \( dx \) to \( du \). Let's differentiate: \( \frac{du}{dx} = \frac{3}{2}x^{1/2} \) or equivalently, \( dx = \frac{2}{3}x^{-1/2}du \).
02

Changing Variables in the Integral

Substitute \( x = u^{2/3} \) and \( dx = \frac{2}{3}x^{-1/2}du \) into the integral:\[ \int \sqrt{\frac{x}{1-x^3}} \, dx = \int \sqrt{\frac{u^{2/3}}{1-(u^{2/3})^3}} \cdot \frac{2}{3} \cdot x^{-1/2} \, du \]. This transforms to \[ \int \sqrt{\frac{u^{2/3}}{1-u^2}} \cdot \frac{2}{3} \cdot (u^{2/3})^{-1/2} \, du \].
03

Simplify the Expression

Now simplify: \( (u^{2/3})^{-1/2} = u^{-1/3} \) so the integral becomes \[ \int \sqrt{\frac{u^{2/3}}{1-u^2}} \cdot \frac{2}{3} \cdot u^{-1/3} \, du = \int \sqrt{\frac{1}{1-u^2}} \cdot \frac{2}{3} \, du \].
04

Trigonometric Substitution

To evaluate \( \int \sqrt{\frac{1}{1-u^2}} \, du \), use the trigonometric substitution \( u = \sin(\theta) \), hence \( du = \cos(\theta) \, d\theta \). This gives us \( \int \sqrt{\frac{1}{1-\sin^2(\theta)}} \cdot \cos(\theta) \, d\theta \).
05

Applying Trigonometric Identity

We know \( 1 - \sin^2(\theta) = \cos^2(\theta) \), so \( \sqrt{\frac{1}{1-u^2}} = \sqrt{\frac{1}{\cos^2(\theta)}} = \frac{1}{\cos(\theta)} \), and the integral becomes \( \int \frac{1}{\cos(\theta)} \cdot \cos(\theta) \, d\theta = \int 1 \, d\theta \).
06

Solve the Integral

The integral \( \int 1 \, d\theta = \theta + C \), where \( C \) is the constant of integration.
07

Back-substitute for Original Variable

Since \( \theta = \arcsin(u) \) and \( u = x^{3/2} \), substitute back to get \( \theta = \arcsin(x^{3/2}) \). Thus the solution to the integral is \( \arcsin(x^{3/2}) + C \).

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Key Concepts

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

Substitution Method
The substitution method is a powerful tool used in integration that simplifies an integral by changing the variables. This technique often makes it easier to evaluate complex integrals.
To apply substitution, you choose a new variable to replace an expression in the integral. This involves differentiating your substitute variable to find the corresponding differential.
In our example, we used the given hint: let \( u = x^{3/2} \). This choice of substitution simplifies the radical expression inside the integral.
By differentiating, we find the relation between \( du \) and \( dx \). These new expressions are then substituted back into the integral, simplifying it and making it more manageable. This process transforms a difficult integrand into a simpler form, paving the way to solve the integral more easily.
Trigonometric Substitution
Trigonometric substitution is another invaluable technique in evaluating integrals, especially those containing radical expressions like \( \sqrt{1-x^2} \). This method involves substituting trigonometric functions for algebraic expressions.
In the example provided, after simplifying the integral using regular substitution, we reach an expression involving \( \sqrt{1-u^2} \). We introduce the trigonometric substitution \( u = \sin(\theta) \), which simplifies the expression.
This substitution is advantageous because trigonometric identities help simplify expressions like \( 1 - \sin^2(\theta) \).
By converting the problem into trigonometric terms, solving the integral becomes straightforward as it often reduces to a standard form. After evaluating the integral, we back-substitute the original variables.
Trigonometric Identities
Trigonometric identities play a crucial role in simplifying integrals that involve trigonometric functions. These identities are equations that are true for all permissible values of the variables.
A popular identity used in integration is \( 1 - \sin^2(\theta) = \cos^2(\theta) \).
In the step-by-step solution, to simplify the expression \( \sqrt{\frac{1}{1-u^2}} \), the identity transforms it to \( \frac{1}{\cos(\theta)} \). This substitution breaks down to a simpler form \( \int 1 \, d\theta \), highlighting the power of trigonometric identities in simplifying and solving integrals.
Definite and Indefinite Integrals
Understanding the difference between definite and indefinite integrals is essential in calculus.
An indefinite integral represents a family of functions and includes a constant of integration, typically indicated with +C. It provides the antiderivative of a function.
A definite integral, on the other hand, has upper and lower limits and represents the area under a curve between those limits. The outcome is a numerical value rather than a function.
In our problem, the integration resulted in an indefinite integral because we did not evaluate it with specific limits. The solution ends with \( \arcsin(x^{3/2}) + C \), showcasing the general antiderivative of the initial integrand.

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Most popular questions from this chapter

A fair coin is tossed four times and the random variable \(X\) assigns the number of tails that appear in each outcome. $$ \begin{array}{l}{\text { a. Determine the set of possible outcomes. }} \\\ {\text { b. Find the value of } X \text { for each outcome. }} \\ {\text { c. Create a probability bar graph for } X, \text { as in Figure } 8.21 . \text { What }} \\ {\text { is the probability that at least two heads appear in the four }} \\ {\text { tosses of the coin? }}\end{array} $$

Use a CAS to perform the integrations. Evaluate the integrals a. \(\int x \ln x d x \quad\) b. \(\int x^{2} \ln x d x \quad\) c. \(\int x^{3} \ln x d x.\) d. What pattern do you see? Predict the formula for \(\int x^{4} \ln x d x\) and then see if you are correct by evaluating it with a CAS. e. What is the formula for \(\int x^{n} \ln x d x, n \geq 1 ?\) Check your answer using a CAS.

Estimating the value of a convergent improper integral whose domain is infinite a. Show that $$\int_{3}^{\infty} e^{-3 x} d x=\frac{1}{3} e^{-9}<0.000042, $$ and hence that \(\int_{3}^{\infty} e^{-x^{2}} d x<0.000042 .\) Explain why this means that \(\int_{0}^{\infty} e^{-x^{2}} d x\) can be replaced by \(\int_{0}^{3} e^{-x^{2}} d x\) without introducing an error of magnitude greater than \(0.000042 .\) b. Evaluate \(\int_{0}^{3} e^{-x^{2}} d x\) numerically. Estimating the value of a convergent improper integral whose domain is infinite

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Use integration, the Direct Comparison Test, or the Limit Comparison Test to test the integrals for convergence. If more than one method applies, use whatever method you prefer. $$\int_{0}^{\pi / 2} \cot \theta d \theta$$
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