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Magazine Chapter 7: Problem 55
Evaluate the integrals. $$\int \frac{3 d r}{\sqrt{1-4(r-1)^{2}}}$$
Short Answer
Expert verified
\( \int \frac{3 \, dr}{\sqrt{1-4(r-1)^2}} = \frac{3}{2} \arcsin(2(r-1)) + C \).
Step by step solution
01
Identify the type of integral
The integral \( \int \frac{3 \, dr}{\sqrt{1 - 4(r-1)^2}} \) resembles the form \( \int \frac{1}{\sqrt{a^2 - u^2}} \, du \), which is an arcsin function. This type of integral leads to an inverse trigonometric function.
02
Substitution for simplification
To bring the integral into the arcsin form, let's substitute \( u = 2(r-1) \). Thus, \( du = 2 \, dr \), which implies that \( dr = \frac{1}{2} \, du \). Substitute these into the integral.
03
Rewrite the integral with substitution
Replace \( r \) with its expression in terms of \( u \):\[\int \frac{3}{2} \frac{du}{\sqrt{1-u^2}}\]Now, we have replaced the original integral with a simpler form suitable for direct integration.
04
Integrate using arcsin formula
The integral \( \int \frac{1}{\sqrt{1-u^2}} \, du \) results in \( \arcsin(u) + C \). Therefore:\[\int \frac{3}{2} \frac{du}{\sqrt{1-u^2}} = \frac{3}{2} \arcsin(u) + C\]
05
Back substitute to original variable
Recall that we substituted \( u = 2(r-1) \). So, replace \( u \) back with this expression:\[\frac{3}{2} \arcsin(2(r-1)) + C\]This is the final evaluated integral back in terms of \( r \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Trigonometric Substitution
Trigonometric substitution is a technique used in calculus to simplify the integration of expressions involving roots such as \( \sqrt{a^2 - x^2} \). This method often converts complex integrals into more manageable forms, which involve known trigonometric identities.
In the original exercise, the expression \( \sqrt{1 - 4(r-1)^2} \) was converted using a substitution. We identified a match for the form \( \sqrt{a^2 - u^2} \) by setting \( u = 2(r-1) \). By rewriting the variable in a trigonometric context, the integral transforms into an easier form. The substitution usually involves using sine, cosine, or tangent substitutions that align with Pythagorean identities. These are:
In the original exercise, the expression \( \sqrt{1 - 4(r-1)^2} \) was converted using a substitution. We identified a match for the form \( \sqrt{a^2 - u^2} \) by setting \( u = 2(r-1) \). By rewriting the variable in a trigonometric context, the integral transforms into an easier form. The substitution usually involves using sine, cosine, or tangent substitutions that align with Pythagorean identities. These are:
- \( x = a\sin(\theta) \)
- \( x = a\cos(\theta) \)
- \( x = a\tan(\theta) \)
Inverse Trigonometric Functions
Inverse trigonometric functions, such as arcsin, arccos, and arctan, are essential in calculus for integrating and transforming expressions that involve radicals. They reverse the normal trigonometric functions, solving for angles when given a trigonometric ratio.
In the exercise, after applying the substitution \( u = 2(r-1) \), the integral transformed into \( \int \frac{du}{\sqrt{1-u^2}} \), which is solved by recognizing it as a known integration formula after substitution: the arcsine function. Specifically, the formula \( \int \frac{1}{\sqrt{1-x^2}} \,dx = \arcsin(x) + C \) applies.
In the exercise, after applying the substitution \( u = 2(r-1) \), the integral transformed into \( \int \frac{du}{\sqrt{1-u^2}} \), which is solved by recognizing it as a known integration formula after substitution: the arcsine function. Specifically, the formula \( \int \frac{1}{\sqrt{1-x^2}} \,dx = \arcsin(x) + C \) applies.
- This technique is powerful because it turns what looks like a complicated expression into an integration of an easily recognizable form.
- The arcsine result from our integral is then expressed in terms of the original variable, completing the process.
Definite Integrals
Definite integrals represent the signed area under a curve between two specific points. Although the exercise involved an indefinite integral, understanding definite integrals provides a complete perspective on integration techniques.
Definite integrals require evaluating the antiderivative at upper and lower boundaries and subtracting the results \(F(b) - F(a)\). The boundaries provide limits that transform an indefinite integral into a specific numerical result.
Definite integrals require evaluating the antiderivative at upper and lower boundaries and subtracting the results \(F(b) - F(a)\). The boundaries provide limits that transform an indefinite integral into a specific numerical result.
- Applications include finding areas, volumes, and average values within certain intervals.
- When using trigonometric substitution with definite integrals, the boundaries must also transform according to the substitution, ensuring consistent evaluations.
