Problem 47 - Calculus of a Single Variable

Question

Comparing Integration Problems In Exercises \(47-50\) , determine which of the integrals can be found using the basic integration formulas you have studied so far in the text. $$ \begin{array}{l}{\text { (a) } \int \frac{1}{\sqrt{1-x^{2}}} d x} \\ {\text { (b) } \int \frac{x}{\sqrt{1-x^{2}}} d x} \\ {\text { (c) } \int \frac{1}{x \sqrt{1-x^{2}}} d x}\end{array} $$

Step-by-Step Solution

Answer

The integral that can be found using the basic integration formulas studied so far is (a) \(\int \frac{1}{\sqrt{1-x^{2}}} dx = \sin^{-1}(x)\). The integrals (b) and (c) require more advanced techniques like trigonometric substitution to solve.

Step 1: Analyze Integral (a)

This is \(\int \frac{1}{\sqrt{1-x^{2}}} d x\). This formula is the derivative of arcsine, written as \(\frac{d}{dx} \sin^{-1}(x) = \frac{1}{\sqrt{1-x^2}}\). Hence, its integral will be \(\sin^{-1}(x)\).

Step 2: Analyze Integral (b)

This is \(\int \frac{x}{\sqrt{1-x^{2}}} d x\). The standard textbook integration formulas do not generally cover this form. However, the integral can still be computed using a technique known as trigonometric substitution; not a basic integration formula.

Step 3: Analyze Integral (c)

This is \(\int \frac{1}{x \sqrt{1-x^{2}}} d x\). Similar to (b), the standard textbook integration formulas do not generally cover this form. The integral can still be computed, though it requires additional techniques like algebraic manipulation and then trigonometric substitution;

Key Concepts

Basic Integration Techniques

Before diving into complex integration problems, let's revisit the foundational concept of basic integration techniques. Integration is a fundamental part of calculus that is concerned with finding the 'whole' from the parts. It's the reverse operation of differentiation, helping to compute areas, volumes, and the value of a function aggregated over a range.

Some of the basic integration formulas include integrating constants, powers of x (like \( \text{{d}}x^n \)), exponential functions, and trigonometric functions. For example, the integral of a constant \( a \) with respect to \( x \) is \( ax + C \), where \( C \) represents the constant of integration. Additionally, a power rule can be applied when integrating powers of x, leading to \( \int x^n \text{{d}}x = \frac{{x^{n+1}}}{{n+1}} + C \), provided \( n eq -1 \).

Understanding and utilizing these formulas is crucial for students as they are not only common in exams but also form the basis for tackling more complex integrals. Moreover, recognizing when a function matches one of these basic forms is a key skill in integration.

Arcsine Function Integration

Integration involving the arcsine function often appears in calculus problems and understanding how to handle this specific scenario is essential. The arcsine function, denoted by \( \text{{sin}}^{-1}(x) \) or \( \text{{asin}}(x) \), is the inverse function of sine and is particularly important when dealing with integrals of the form \( \int \frac{{1}}{{\text{{sqrt}}(1-x^2)}} \text{{d}}x \).

In the exercise provided, integral (a) \( \int \frac{{1}}{{\text{{sqrt}}(1-x^2)}} \text{{d}}x \) can be directly related to the arcsine function because the derivative of \( \text{{asin}}(x) \) gives the integrand \( \frac{{1}}{{\text{{sqrt}}(1-x^2)}} \). Therefore, the antiderivative, or integration, of this expression is \( \text{{sin}}^{-1}(x) + C \), where \( C \) represents the constant of integration.

Knowing the relationship between the arcsine function and its derivative simplifies the process of integration in such cases, providing a quick and elegant solution to what could otherwise be a challenging integral.

Trigonometric Substitution

Trigonometric substitution is an advanced integration technique that is employed when the integrand includes expressions such as \( \text{{sqrt}}(a^2 - x^2) \), \( \text{{sqrt}}(a^2 + x^2) \), or \( \text{{sqrt}}(x^2 - a^2) \). It involves substituting a trigonometric expression for a variable to simplify the integral and make it easier to solve.

For instance, in the exercise under consideration, for integrals (b) and (c), standard textbook formulas are not sufficient. In these cases, trigonometric substitution can be used. Taking integral (b) \( \int \frac{{x}}{{\text{{sqrt}}(1-x^2)}} \text{{d}}x \), one could substitute \( x = \text{{sin}}(\theta) \), transforming the square root term into a simple trigonometric function which may be integrated using basic trigonometric integrals.

This technique often requires the additional step of finding the bounds of integration in terms of the new variable, which can add to the complexity. Nonetheless, trigonometric substitution remains a powerful tool in the integration toolbox. A solid understanding of trigonometric identities and inverse functions is needed to effectively use trigonometric substitution, reinforcing its status as an advanced technique in calculus.