Problem 55 - Applied Calculus

Question

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

Step-by-Step Solution

Answer

The integral is \( -\ln\left| \frac{1}{x} + \frac{\sqrt{1-x^2}}{x} \right| + \sqrt{1-x^2} + C \).

Step 1: Substitution Preparation

Notice the form \( \sqrt{1-x^2} \). This is reminiscent of the trigonometric identity \( \sin^2\theta + \cos^2\theta = 1 \). We can make the substitution \( x = \sin\theta \), which gives \( dx = \cos\theta \, d\theta \). The integrand becomes \( \sqrt{1-\sin^2\theta} \) which simplifies to \( \cos\theta \).

Step 2: Perform the Substitution

Substitute \( x = \sin\theta \) and \( dx = \cos\theta \,d\theta \) into the integral: \[ \int \frac{\sqrt{1-x^2}}{x} \, dx = \int \frac{\cos\theta}{\sin\theta} \cdot \cos\theta \, d\theta = \int \frac{\cos^2\theta}{\sin\theta} \, d\theta. \]

Step 3: Rewrite the Integrand

Rewrite \( \frac{\cos^2\theta}{\sin\theta} \) as \( \frac{1-\sin^2\theta}{\sin\theta} \). This breaks down into two separate integrals: \[ \int \left( \frac{1}{\sin\theta} - \sin\theta \right) \, d\theta = \int \csc\theta \, d\theta - \int \sin\theta \, d\theta. \]

Step 4: Evaluate the Integrals

The integral \( \int \csc\theta \, d\theta \) is a standard result given by \( -\ln|\csc\theta + \cot\theta| + C_1 \). The integral \( \int \sin\theta \, d\theta \) is \( -\cos\theta + C_2 \). Combining these results, the integral becomes: \[ -\ln|\csc\theta + \cot\theta| + \cos\theta + C. \]

Step 5: Back-Substitute \( \theta \)

Recall that \( x = \sin\theta \), thus \( \csc\theta = \frac{1}{\sin\theta} = \frac{1}{x} \) and \( \cot\theta = \frac{\cos\theta}{\sin\theta} = \frac{\sqrt{1-x^2}}{x} \). Therefore, back-substitute \( \theta \) to get the solution in terms of \( x \): \[ \int \frac{\sqrt{1-x^2}}{x} \, dx = -\ln\left| \frac{1}{x} + \frac{\sqrt{1-x^2}}{x} \right| + \sqrt{1-x^2} + C. \]

Key Concepts

Trigonometric Substitution

Trigonometric substitution is a clever technique used in calculus to simplify integrals that involve square roots. When faced with expressions like \( \sqrt{1-x^2} \), it taps into the Pythagorean identity: \( \sin^2\theta + \cos^2\theta = 1 \). This suggests that substituting \( x = \sin\theta \) can simplify the integral significantly.

Here's why it works:

• By substituting \( x = \sin\theta \), you can change a complicated integral into one that involves trigonometric functions, which are often easier to integrate.
• With this substitution, \( dx \) becomes \( \cos\theta \, d\theta \), which further aligns the structure of the integral for simplification.
• After substitution, expressions like \( \sqrt{1-x^2} \) become simpler trigonometric terms such as \( \cos\theta \).

Once the substitution is complete, you solve the integral in terms of \( \theta \) and then back-substitute the original variable, \( x \). This technique is especially useful for integrals featuring square root expressions of the form \( \sqrt{a^2 - x^2} \).

Definite and Indefinite Integrals

Understanding definite and indefinite integrals is crucial in calculus. An indefinite integral represents a family of functions and includes a constant of integration \( C \). In contrast, a definite integral calculates the net area under a curve within given limits.

• Indefinite integrals are general solutions to the integration problem. They do not evaluate a specific area but rather provide a formula that can be used for any specific interval.
• In calculations, indefinite integrals are notated as \( \int f(x) \, dx = F(x) + C \), where \( F(x) \) is the antiderivative of \( f(x) \).

Definite integrals, however, are not involved in this particular problem since we did not evaluate the integral over a specific interval, just found a general solution with the indefinite integral approach. The knowledge of both types of integrals is essential for solving a wide range of calculus problems because it allows you to handle various scenarios, whether you need a general solution or a specific numerical result.

Calculus Problem Solving

Solving calculus problems often requires an understanding of multiple techniques, such as trigonometric substitution, as seen in this example. Each integral presents unique challenges, so having a flexible approach is key.

• Identify the type of problem: Recognize the form of the function under the integral to choose an appropriate technique, like substitution or integration by parts.
• Apply the technique: For complicated integrals, don't hesitate to utilize established identities and substitutions, as they can simplify the problem significantly.
• Simplify wherever possible: Breaking the integral into simpler parts can make computation easier, as demonstrated by rewriting terms using trigonometric identities.

The goal is to turn a complicated integral into a simpler problem, one that matches forms we know how to solve. This step-by-step approach not only makes integration more manageable but also enhances understanding of the underlying principles of calculus. As you practice, you'll become more efficient and confident in handling different types of calculus problems.