Problem 59 - Thomas Calculus

Question

Evaluate the integrals in Exercises \(47-70\) $$ \int \frac{3 d r}{\sqrt{1-4(r-1)^{2}}} $$

Step-by-Step Solution

Answer

Evaluate to \( \frac{3}{2} \sin^{-1}(2(r-1)) + C \).

Step 1: Simplify the Expression

Let's first simplify the expression under the square root: \(1 - 4(r-1)^2\). We have: \[ \sqrt{1 - 4(r-1)^2} \]. This can be rewritten as \( \sqrt{1 - 4((r-1))^2 + 4} - 4 \). This simplification helps to prepare the expression for further manipulation in terms of trigonometric substitution.

Step 2: Substitute Variable for Simplification

We'll use a trigonometric substitution to solve the integral. Set \( u = r - 1 \), so \( d u = d r \). The integral becomes: \[ \int \frac{3 \, d u}{\sqrt{1 - 4 u^2}} \].

Step 3: Use Trigonometric Substitution

Use the substitution \( u = \frac{1}{2} \sin \theta \), transforming the integral into: \[ \int \frac{3 \, d(\frac{1}{2} \sin \theta)}{\sqrt{1 - 4(\frac{1}{2} \sin \theta)^2}} \]. Calculate \( d u = \frac{1}{2} \cos \theta \cdot d \theta \). Thus, our integral converts to: \[ \int \frac{3 \cdot \frac{1}{2} \cos(\theta) \, d \theta}{\sqrt{1 - \sin^2(\theta)}} = \int \frac{3}{2} \cos(\theta) \, d \theta \].

Step 4: Simplify the Integral with Trigonometric Identities

Since \( \sqrt{1 - \sin^2(\theta)} = \cos(\theta)\), the terms simplify to: \[ \int \frac{3}{2} \, d \theta \]. This integral can now be easily resolved.

Step 5: Integrate the Remaining Expression

Integrating \( \frac{3}{2} \, d \theta \) results in: \[ \frac{3}{2} \theta + C \], where \( C \) is our constant of integration.

Step 6: Back Substitute to Original Variables

Recall \( \theta = \sin^{-1}(2u) \), and \( u = r - 1 \). Therefore, \( \theta = \sin^{-1}(2(r-1)) \). Replacing \( \theta \) back into the expression gives us: \[ \frac{3}{2} \sin^{-1}(2(r-1)) + C \].

Step 7: Final Step: Write the Solution

The evaluated integral is: \[ \frac{3}{2} \sin^{-1}(2(r-1)) + C \].

Key Concepts

Trigonometric Substitution

Trigonometric substitution is a powerful technique used in calculus to evaluate integrals, especially those involving square roots. The idea is to simplify the integral by replacing a variable with a trigonometric function.
This often allows us to reduce complex expressions to more manageable forms. In the given exercise, the term under the square root, \(1 - 4(r-1)^2\), suggests the potential for trigonometric substitution.

• The substitution used here is \( u = \frac{1}{2} \sin \theta \).
• This substitution turns the problem into an integral involving basic trigonometric functions.

This is effective because trigonometric identities provide simpler expressions for complex roots. The substitution exploits the identity \(1 - \sin^2 \theta = \cos^2 \theta\).
Converting the variable this way turns the square root expression into \( \cos \theta \), simplifying the integral to a basic form. This substitution makes solving the integral more straightforward, allowing us to apply simple trigonometric identities to resolve the problem.

Definite and Indefinite Integrals

Integrals can be classified as either definite or indefinite. An indefinite integral represents a family of functions and includes a constant of integration, denoted \(C\). In our exercise, we are dealing with an indefinite integral.
The goal with indefinite integrals is to find the antiderivative of the function.

• In an indefinite integral, unlike definite integrals, there are no upper or lower limits of integration.
• Thus, the solution represents a general form of a function, not a specific numerical value.

In this problem, after substitution and simplification, we end up with an integral of the form \( \int \frac{3}{2} \cos(\theta) \, d \theta \). Solving this yields the result \( \frac{3}{2} \theta + C \).
Where \(C\) represents the constant of integration. The last step involves back-substituting to the original variable \(r\), finalizing the integral in terms of the original function.
This results in \( \frac{3}{2} \sin^{-1}(2(r-1)) + C \), which gives us the indefinite integral of the function.

Trigonometric Identities

Trigonometric identities are equations that involve trigonometric functions and are true for every value of the occurring variables. They are extremely useful in simplifying complex calculus problems.
In this exercise, employing these identities makes it easier to handle the expression under the root.

• The primary identity used here is \( \cos^2 \theta + \sin^2 \theta = 1 \).
• This converts the square root of \(1 - \sin^2 \theta\) directly to \( \cos \theta \).

Using this identity, the troublesome square root in the denominator simplifies down to \( \cos \theta \). This eliminates the complexities involved in integrating functions with radical expressions.
By incorporating such identities, we turn the integration process into one involving simple trigonometric functions, making it manageable and straightforward to solve.
This is a typical strategy in calculus to handle problems involving square roots and powers, highlighting the power and elegance of using trigonometric substitutions & identities together.