Problem 54 - Applied Calculus

Question

Find each integral by whatever means are necessary (either substitution or tables). $$ \int \frac{x}{\sqrt{1-x^{2}}} d x $$

Step-by-Step Solution

Answer

The integral solution is \(-\sqrt{1-x^2} + C\).

Step 1: Recognize the Form

Notice that the form of the integral resembles a common derivative rule. Recall that the derivative of \( \arcsin{x} \) is \( \frac{1}{\sqrt{1-x^2}} \).

Step 2: Set Up for Substitution

We recognize this integral might involve substitution. We set \( u = 1 - x^2 \), then \( du = -2x \, dx \) or \( -\frac{1}{2} du = x \, dx \).

Step 3: Rewrite Integral Using Substitution

Substitute \( u = 1 - x^2 \) into the integral. The integral becomes \(-\frac{1}{2} \int \frac{1}{\sqrt{u}} du\).

Step 4: Solve the Integral

The integral \( \int \frac{1}{\sqrt{u}} du \) simplifies to \( 2\sqrt{u} + C \). Using the substitution back, the integral becomes \(-\frac{1}{2}(2\sqrt{1-x^2}) + C\).

Step 5: Simplify the Result

Simplify the expression: \(-\sqrt{1-x^2} + C\). This is the antiderivative of the original function.

Key Concepts

Substitution Method

The substitution method is a technique used to simplify the integration process by changing variables. It works particularly well when dealing with functions that can be transformed into a more familiar format. Imagine trying to solve a puzzle – instead of approaching the problem head-on, substitution allows you to break it into simpler pieces.

To use the substitution method, follow these steps:

• Identify a part of the integral that can be replaced with a single variable, often denoted as \(u\).
• Express the original integral in terms of \(u\), and find the differential \(du\) in relation to the original variable.
• Rewrite the integral in terms of \(u\) and solve the simpler form.
• After integration, substitute back the original variable to express the result in the original terms.

In the given problem, setting \(u = 1 - x^2\) transforms the integral into one involving \(u\), turning a challenging problem into something more manageable. This method shows its effectiveness, especially when functions have derivatives similar to known forms.

Definite and Indefinite Integrals

Understanding the difference between definite and indefinite integrals is crucial in solving integration problems. An indefinite integral, like the one in our problem, represents a family of functions. It includes a constant of integration, usually denoted by \(C\), as it can vary by a constant.

Indefinite integrals:

• Do not have specified limits; they represent the general antiderivative of a function.
• Require adding a constant \(C\) because integration is the reverse of differentiation.

On the other hand, definite integrals compute the area under a curve between specified limits:

• They have upper and lower bounds to confine the calculation to a specific interval.
• They result in a specific numerical value rather than a function.

In this exercise, since there are no specified limits in the given integral \(\int \frac{x}{\sqrt{1-x^2}} dx\), it is an indefinite integral. The solution leads to \(-\sqrt{1-x^2} + C\), with \(C\) indicating this general solution.

Trigonometric Substitution

Trigonometric substitution is a powerful technique in calculus used to solve integrals involving square roots. It involves substituting trigonometric functions for expressions involving radicals, exploiting the Pythagorean identities to simplify the integral.

When to use trigonometric substitution:

• Whenever you encounter integrals with expressions like \(\sqrt{a^2 - x^2}\), \(\sqrt{a^2 + x^2}\), or \(\sqrt{x^2 - a^2}\).
• These expressions are perfect candidates for replacing with trigonometric identities.

For example, for \(\sqrt{1-x^2}\), substituting \(x = \sin(\theta)\) transforms it into \(\cos(\theta)\) using the identity \(\sin^2(\theta) + \cos^2(\theta) = 1\). This approach leverages the circular nature of trigonometric functions to transform complex radicals into something easier to manage.

In this particular problem, identifying that the expression resembles a trigonometric form makes solving it more intuitive when aligned with other methods such as simple substitution.