Problem 51 - Thomas Calculus

Question

Evaluate the integrals in Exercises 51 and 52. $$ \int \frac{(2 r-1) \cos \sqrt{3(2 r-1)^{2}+6}}{\sqrt{3(2 r-1)^{2}+6}} d r $$

Step-by-Step Solution

Answer

\( \sin(\sqrt{3(2r - 1)^2 + 6}) + C \)

Step 1: Identify the Integration Technique

The given integral presents a functional form suggesting the use of substitution. Because of the complex expression inside the cosine function and the square root in the denominator, substitution will simplify the integration process.

Step 2: Substitution

Let's make the substitution where \( u = \sqrt{3(2r - 1)^2 + 6} \). Then, differentiate both sides with respect to \( r \):\[ du = \frac{d}{dr} \sqrt{3(2r - 1)^2 + 6} \, dr \]

Step 3: Differentiate the Substituted Expression

Compute the derivative of \( u \) with respect to \( r \):\[ du/dr = \frac{1}{2\sqrt{3(2r - 1)^2 + 6}} \cdot 6(2r - 1) \cdot 2 \]Multiply both sides by \( dr \) to get:\( du = \frac{(2r - 1)}{\sqrt{3(2r - 1)^2 + 6}} \, dr \)

Step 4: Substitute \( du \) in the Integral

Replace the integrand and differential \( dr \) using our substitution:\[ \int \frac{(2r - 1) \cos(\sqrt{3(2r - 1)^2 + 6})}{\sqrt{3(2r - 1)^2 + 6}} \, dr = \int \cos(u) \, du \]

Step 5: Integrate with Respect to \( u \)

Now, the integral simplifies to:\[ \int \cos(u) \, du = \sin(u) + C \]Where \( C \) is the constant of integration.

Step 6: Substitute Back to Original Variable

Go back to the variable \( r \) by substituting \( u = \sqrt{3(2r - 1)^2 + 6} \) back:\[ \int \cos(u) \, du = \sin(\sqrt{3(2r - 1)^2 + 6}) + C \]

Key Concepts

Substitution Method

The substitution method is a powerful technique used to simplify complex integrals, making them easier to evaluate. This approach is especially useful when dealing with a composite function, like in the given exercise. The main idea is to turn the original variable into a new variable, which often simplifies the integrand. The steps to apply substitution are straightforward:

• Select a substitution that simplifies the expression. Look for expressions that appear repeatedly or complex parts inside a function, like square roots or trigonometric expressions.
• Express the differential of the new variable in terms of the original variable. This involves differentiating the substitution equation.
• Replace the original variable and differential in the integral with the new variable and its differential. This step usually results in a much simpler integral.

In the problem, using substitution transformed a complex expression into a simple integral, allowing easier integration of the trigonometric function.

Definite and Indefinite Integrals

Integrals can be categorized as definite or indefinite, and both are crucial in calculus. An indefinite integral represents a family of functions and includes an arbitrary constant, denoted as + C:

• An indefinite integral is used when the antiderivative of a function is determined without specific bounds. The solution provides a general form, covering all possible functions that could have differentiated to form the integrand.
• A definite integral, on the other hand, evaluates the area under a curve between two limits. It results in a numeric value and does not include the constant of integration.

In our exercise, we deal with an indefinite integral as we simplify the expression step by step and incorporate the constant of integration after finding the antiderivative.

Trigonometric Integration

Trigonometric integration involves integrating functions that include trigonometric expressions. Common trigonometric functions like sine and cosine frequently appear in integrals, requiring specific techniques:

• Basic integrals: Knowing the basic antiderivatives of trigonometric functions is essential. For instance, the integral of \(\cos{x}\) is \(\sin{x} + C\).
• Trigonometric identities: These can simplify expressions involving powers or products of trig functions.
• Specific substitutions: Often used for integrals involving expressions like \(a^2 - x^2\) or \(x^2 - a^2\), transforming them into trigonometric forms that are more straightforward to integrate.

In this exercise, we used the basic integral of cosine to find the antiderivative, resulting in a simple expression. This underscores the utility of trigonometric integration when dealing with seemingly complex functions that involve trigonometric components.