Chapter 5: Problem 282
In the following exercises, use a suitable change of variables to determine the indefinite integral. $$ \int \frac{x^{3}}{\sqrt{1-x^{2}}} d x $$
Short Answer
Expert verified
\( \int \frac{x^3}{\sqrt{1-x^2}} \, dx = \frac{1}{3}(1-x^2)^{3/2} - \sqrt{1-x^2} + C \)
Step by step solution
01
Identify the Substitution
Given the integral \( \int \frac{x^3}{\sqrt{1-x^2}} \, dx \), notice the presence of \( \sqrt{1-x^2} \). A suitable substitution is \( x = \sin(\theta) \), because \( 1 - \sin^2(\theta) = \cos^2(\theta) \) simplifies the square root.
02
Differentiate to Find dx
Differentiate \( x = \sin(\theta) \) to find \( dx \). We get \( dx = \cos(\theta) \, d\theta \). Substitute \( dx \) in the integral.
03
Simplify the Integral
Substitute \( x = \sin(\theta) \) and \( dx = \cos(\theta) \, d\theta \) into the integral:\[\int \frac{(\sin(\theta))^3}{\sqrt{1-\sin^2(\theta)}} \cos(\theta) \, d\theta = \int \frac{\sin^3(\theta) \cos(\theta)}{\cos(\theta)} \, d\theta = \int \sin^3(\theta) \, d\theta\]
04
Simplify Further Using Trigonometric Identity
Use the identity \( \sin^2(\theta) = 1 - \cos^2(\theta) \) to express \( \sin^3(\theta) \) as \( \sin(\theta) (1 - \cos^2(\theta)) \). The integral becomes:\[ \int \sin(\theta) - \sin(\theta) \cos^2(\theta) \, d\theta = \int \sin(\theta) \, d\theta - \int \sin(\theta) \cos^2(\theta) \, d\theta\]
05
Integrate the Parts
Integrate each part separately:1. \( \int \sin(\theta) \, d\theta = -\cos(\theta) + C_1 \)2. For \( \int \sin(\theta) \cos^2(\theta) \, d\theta \), use substitution \( u = \cos(\theta), \, du = -\sin(\theta) \, d\theta \): \[ \int -u^2 \, du = -\frac{u^3}{3} + C_2 = -\frac{\cos^3(\theta)}{3} + C_2 \]
06
Combine and Back-Substitute
Combine both integrals:\[-\cos(\theta) + \frac{\cos^3(\theta)}{3} + C\]Now, back-substitute \( \theta = \arcsin(x) \):\[-\sqrt{1-x^2} + \frac{(1-x^2)^{3/2}}{3} + C\]
07
Simplify and Present the Final Solution
The simplified form of the integral is:\[\frac{1}{3}(1-x^2)^{3/2} - \sqrt{1-x^2} + C\]This is the solution to the indefinite integral.
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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 technique used in calculus to simplify the process of finding integrals. It involves replacing the variable of integration with a new variable that might make the integral easier to evaluate.
In our exercise, we deal with the indefinite integral \( \int \frac{x^3}{\sqrt{1-x^2}} \ dx \). Here, it's helpful to use a trigonometric substitution. A suitable variable change can transform the complex expression into something more manageable.
In our exercise, we deal with the indefinite integral \( \int \frac{x^3}{\sqrt{1-x^2}} \ dx \). Here, it's helpful to use a trigonometric substitution. A suitable variable change can transform the complex expression into something more manageable.
- Identify an expression that is complicated, such as a compound function or a square root.
- Choose a substitution that simplifies this expression. In this case, \( x = \sin(\theta) \) is ideal because it transforms \( \sqrt{1-x^2} \) into \( \cos(\theta) \).
- Replace the original variable with the new one and remember to substitute \( dx \) as well (here, \( dx = \cos(\theta) \, d\theta \)).
Trigonometric Substitution
Trigonometric substitution is especially powerful for integrals involving square roots of the form \( \sqrt{a^2 - x^2} \), \( \sqrt{a^2 + x^2} \), or \( \sqrt{x^2 - a^2} \). By substituting the variable \( x \) with a trigonometric function, we make use of Pythagorean identities.
In the exercise, we chose \( x = \sin(\theta) \) because \( 1-x^2 \) becomes \( \cos^2(\theta) \), simplifying the square root easily. This is suitable since:
In the exercise, we chose \( x = \sin(\theta) \) because \( 1-x^2 \) becomes \( \cos^2(\theta) \), simplifying the square root easily. This is suitable since:
- The identity \( 1 - \sin^2(\theta) = \cos^2(\theta) \) applies, reducing the square root in the denominator.
- \( \sin(\theta) \) and \( \cos(\theta) \) are well-understood functions with known derivative and integrals, making calculations straightforward.
Integration by Parts
Integration by parts is a method derived from the product rule of differentiation and is used to integrate the product of two functions. The formula is given by:\[\int u \, dv = uv - \int v \, du.\]
In our problem, integration by parts is used indirectly when breaking down \( \int \sin(\theta) \cos^2(\theta) \, d\theta \). Instead, substituting \( u = \cos(\theta) \) converts this integral into a polynomial form \( \int -u^2 \, du \), which can be easily integrated using basic rules.
The idea is to decompose a complex integral into simpler parts:
In our problem, integration by parts is used indirectly when breaking down \( \int \sin(\theta) \cos^2(\theta) \, d\theta \). Instead, substituting \( u = \cos(\theta) \) converts this integral into a polynomial form \( \int -u^2 \, du \), which can be easily integrated using basic rules.
The idea is to decompose a complex integral into simpler parts:
- Choose \( u \) and \( dv \) from the integral such that \( dv \) is easily integrable.
- Calculate \( du \) by differentiating \( u \) and find \( v \) by integrating \( dv \).
- Apply the integration by parts formula to simplify the integral.