Problem 24 - Thomas Calculus

Question

Evaluate the integrals in Exercises \(1-28\). $$ \int \frac{\left(1-x^{2}\right)^{1 / 2}}{x^{4}} d x $$

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{\sqrt{1-x^2}}{x} + \frac{1}{3}\left( \frac{\sqrt{1-x^2}}{x} \right)^3 + C \).

Step 1: Use Substitution

To simplify the integral, we perform the substitution \( x = rac{1}{t} \) which implies \( dx = -rac{1}{t^2} dt \). The function inside the integral becomes \( 1 - x^2 = 1 - rac{1}{t^2} = \frac{t^2 - 1}{t^2} \). Our integral now becomes \( \ \int \frac{(1-x^2)^{1/2}}{x^4} \, dx = \int \frac{\left(\frac{t^2-1}{t^2}\right)^{1/2}}{\left(\frac{1}{t}\right)^4} \left( -\frac{1}{t^2} \right) dt = \int \frac{t(t^2-1)^{1/2}}{1} \, dt \, \ \).

Step 2: Simplify the Integral

The substitution leads us to the simplified integral \( \int t(t^2 - 1)^{1/2} dt \). This looks like a trigonometric substitution, so let's consider \( t = \sec(\theta) \), then \( dt = \sec(\theta)\tan(\theta) \, d\theta \). The expression \( (t^2 - 1)^{1/2} = (\sec^2(\theta) - 1)^{1/2} = \tan(\theta) \), simplifying our integral to \( \int \sec(\theta)\tan^2(\theta)\tan(\theta)\sec(\theta) \, d\theta \), which reduces to \( \int \sec^2(\theta)\tan^2(\theta) \, d\theta \).

Step 3: Evaluate the Trigonometric Integral

To solve \( \int \sec^2(\theta) \tan^2(\theta) \, d\theta \), we recognize \( \tan^2(\theta) = \sec^2(\theta) - 1 \). Consequently, the integral becomes \( \int (\sec^4(\theta) - \sec^2(\theta)) \, d\theta \). Split this into two separate integrals: \( \int \sec^4(\theta) \, d\theta - \int \sec^2(\theta) \, d\theta \). The second integral is straightforward \( \int \sec^2(\theta) \, d\theta = \tan(\theta) + C \), and the first integral can be converted to a lower power using the identity \( \sec^2(\theta) = 1 + \tan^2(\theta) \).

Step 4: Simplify and Solve Remaining Integral

Returning to \( \int \sec^4(\theta) \, d\theta = \int (1 + \tan^2(\theta)) \sec^2(\theta) \, d\theta \), this can be split into \( \int \sec^2(\theta) \, d\theta + \int \tan^2(\theta)\sec^2(\theta) \, d\theta \), where the second integral \( \int \tan^2(\theta)\sec^2(\theta) \, d\theta \) can be further simplified since \( u = \tan(\theta) \), so \( du = \sec^2(\theta) \, d\theta \), and \( \int \tan^2(\theta)\sec^2(\theta) \, d\theta = \int u^2 \, du = \frac{u^3}{3} + C \).

Step 5: Back-Substitute and Finalize Result

For \( \int \tan^2(\theta)\sec^2(\theta) \, d\theta \), we revert back to \( u = \tan(\theta) \), meaning we end with \( \frac{u^3}{3} + C = \frac{\tan^3(\theta)}{3} + C \). Our complete integral becomes \( \tan(\theta) + \frac{\tan^3(\theta)}{3} + C \). Returning to original variables, since \( t = \sec(\theta) \) and \( x = 1/t \), \( \tan(\theta) = \sqrt{t^2 - 1} \), then \( \tan(\theta) = \sqrt{\frac{1}{x^2} - 1} = \frac{\sqrt{1-x^2}}{x} \). Thus, the integral evaluates to \( \frac{\sqrt{1-x^2}}{x} + \frac{1}{3}\left( \frac{\sqrt{1-x^2}}{x} \right)^3 + C \).

Key Concepts

Integration Techniques

Integration is a fundamental concept in calculus used to find areas under curves, among other applications. Various integration techniques exist to tackle different types of integrals.
Depending on the complexity and the form of the integrand (the function being integrated), certain techniques work best.

• Substitution: This technique involves replacing a part of the integral with a simpler variable to make the integration process more straightforward.
• Trigonometric Substitution: Useful when the integrand involves expressions like \(\sqrt{a^2 - x^2}\), \(\sqrt{a^2 + x^2}\), or \(\sqrt{x^2 - a^2}\). Involving trigonometric identities helps simplify these expressions.
• Integration by Parts: Based on the product rule for differentiation, this technique is useful for integrands that are products of two functions.

The choice of technique depends on recognizing the best approach to simplify and evaluate the integral.

Trigonometric Substitution

Trigonometric substitution is a method that can simplify integration by replacing algebraic expressions with trigonometric ones. This technique leverages trigonometric identities to simplify the integration process.
It is particularly useful when dealing with roots, such as \(\sqrt{1-x^2}\).
To apply this method:

• Identify the form of the integrand and make an appropriate substitution. For example, use \(x = \sin(\theta)\) if you have \(\sqrt{1-x^2}\).
• Convert the integral entirely in terms of \(\theta\), using \(dx = \cos(\theta)d\theta\) for \(x = \sin(\theta)\).
• Integrate using trigonometric identities. For example, \(\sin^2(\theta) + \cos^2(\theta) = 1\).
• Finally, convert back the result to the original variable using the inverse trigonometric function.

By converting complicated roots into trigonometric terms, this method can significantly simplify integration tasks.

Definite and Indefinite Integrals

In calculus, integrals are categorized as definite or indefinite integrals, depending on whether they have limits of integration.
Understanding the difference is crucial in interpreting results.

• Indefinite Integrals: Represent the general form of antiderivatives of a function. It's like the reverse process of differentiation. An indefinite integral appears without set bounds and includes a constant of integration, \(C\), for instance, \(\int f(x) \, dx = F(x) + C\).
• Definite Integrals: Calculated between two limits, they represent the net area under the curve of the function over a specified interval. Written as \(\int_{a}^{b} f(x) \, dx\), where \(a\) and \(b\) are the limits, they yield a specific value.

Though definite integrals result in a numerical value, indefinite integrals yield function expressions.

Integration by Parts

Integration by parts is a technique derived from the product rule of differentiation. It's particularly useful for functions expressed as a product of two simpler functions.
The formula for integration by parts is:\[ \int u \, dv = uv - \int v \, du \]To use integration by parts:

• Identify: Break the integrand into two parts – one being \(u\) (easily differentiable) and the other \(dv\) (easily integrable).
• Differentiate and Integrate: Differentiate \(u\) to find \(du\), and integrate \(dv\) to find \(v\).
• Substitute: Use the formula to find the integral and solve.

Sometimes, you may need to apply integration by parts more than once or combine it with other techniques to solve complex integrals successfully.