Problem 28 - Thomas Calculus

Question

Evaluate the integrals in Exercises \(1-34\) without using tables. $$ \int_{0}^{1} \frac{4 r d r}{\sqrt{1-r^{4}}} $$

Step-by-Step Solution

Answer

The integral evaluates to 2.

Step 1: Recognize the Integral Type

The integral \( \int_{0}^{1} \frac{4r \, dr}{\sqrt{1 - r^4}} \) resembles a trigonometric substitution problem due to the square root in the denominator.

Step 2: Substitute for Simplification

Use the substitution \( r = \sin^{1/2}(\theta) \). This implies \( dr = \frac{1}{2} \cos(\theta) \sin^{-1/2}(\theta) \, d\theta \). The integral becomes \( \int \frac{4 \sin^{1/2}(\theta) \cdot \frac{1}{2} \cos(\theta) \sin^{-1/2}(\theta) \, d\theta}{\sqrt{1-(\sin(\theta))^2}} \).

Step 3: Simplify the Expression

With the energy identity \( \sqrt{1 - \sin^2(\theta)} = \cos(\theta) \), the integral simplifies to \( 2 \int \frac{\sin(\theta)\cos(\theta)\,d\theta}{\cos(\theta)} \), which further simplifies to \( 2 \int \sin(\theta)\, d\theta \).

Step 4: Integrate the Simplified Function

The integral \( \int \sin(\theta) \, d\theta \) is \( -\cos(\theta) + C \). Thus, \( 2 \int \sin(\theta) \, d\theta \) becomes \( -2\cos(\theta) + C \).

Step 5: Re-substitute and Evaluate the Limits

Since \( r = \sin^{1/2}(\theta) \), we need to find \( \theta \) for the limits \( r = 0 \) to \( r = 1 \). At \( r = 0 \), \( \theta = 0 \). At \( r = 1 \), \( \theta = \frac{\pi}{2} \). Substitute to get \( -2\cos\left( \frac{\pi}{2} \right) + 2\cos(0) = 0 - 2(-1) = 2 \).

Step 6: Check the Solution

Verify correctness by checking the integral computation and the substitutions. Everything is consistent, and previous calculations confirm that the result \( 2 \) is accurate.

Key Concepts

Trigonometric Substitution

Trigonometric substitution is an integration technique that involves substituting a trigonometric function for a variable to simplify the problem. This method is particularly useful when dealing with integrals that involve square roots of expressions, such as \( \sqrt{1 - x^2} \), \( \sqrt{a^2 + x^2} \), or \( \sqrt{x^2 - a^2} \). In the context of the example problem, you notice a square root in the denominator, which indicates that a trigonometric substitution might simplify the integral.

Generally, when you have an integral involving \( \sqrt{a^2 - x^2} \), substituting \( x = a \sin\theta \) can simplify the problem. Then, \( \sqrt{a^2 - x^2} = a \cos\theta \). For \( \sqrt{a^2 + x^2} \), the substitution \( x = a \tan\theta \) is useful, and for \( \sqrt{x^2 - a^2} \), \( x = a \sec\theta \) works.

In the original integral provided, substituting \( r = \sin^{1/2}(\theta) \) transformed it into a more manageable form by turning the irrational part into a trigonometric identity, facilitating easier integration.

Definite Integrals

Definite integrals provide a way to calculate the net area under a curve within certain limits, often from \( a \) to \( b \). This is different from indefinite integrals, which represent a family of functions and include an arbitrary constant \( C \).

In the solved exercise, the problem involves calculating the definite integral \( \int_{0}^{1} \frac{4r \, dr}{\sqrt{1 - r^4}} \). This requires you to not only integrate the expression but also evaluate it at the specific upper and lower limits. The substitution connects these limits with the new variable.

Substituting \( r = \sin^{1/2}(\theta) \), you find that the limits change from \( r = 0 \) to \( r = 1 \) on the integral to \( \theta = 0 \) and \( \theta = \frac{\pi}{2} \).

Once you've solved the integral, the last step is substituting back to calculate the net area within the given bounds, which provides the definite integral's result.

Integration Techniques

Integration techniques are strategies or methods for finding the antiderivative or integrating complex functions. Trigonometric substitution is just one of these techniques.

In general, you might find various methods helpful, such as:

• Substitution Method: This technique involves changing the variable to simplify the integral, similar to what was done in trigonometric substitution.
• Integration by Parts: Useful for integrals involving products of functions, leveraging the formula derived from the product rule for derivatives.
• Partial Fraction Decomposition: Often, algebraic manipulation in integrals with rational functions can simplify the process, turning complex rational expressions into simpler ones that can be integrated directly.
• Numerical Integration: For cases when analytical methods are infeasible, numerical techniques like Simpson's Rule or the Trapezoidal Rule can approximate the value of the integral.

In the specific example, the use of trigonometric identities and simplifications brought the integral down to a basic trigonometric form, \( 2 \int \sin(\theta)\, d\theta \), making it easier to integrate and understand.