Problem 55 - Calculus Single Variable

Question

Calculate the given integral. $$ \int \frac{\sqrt{1+x^{2}}}{x^{2}} d x $$

Step-by-Step Solution

Answer

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

Step 1: Identify the Type of Integral

The integral \( \int \frac{\sqrt{1+x^{2}}}{x^{2}} \, dx \) involves a radical function divided by a polynomial. It's suitable for substitution or trigonometric identities. We need to simplify the expression for easier integration.

Step 2: Use Trigonometric Substitution

Use the substitution \( x = \tan(\theta) \) which implies \( dx = \sec^2(\theta) \, d\theta \). The functions transform to \( \sqrt{1+x^2} = \sec(\theta) \) and \( x^2 = \tan^2(\theta) \). This changes the integral accordingly: \( \int \frac{\sec(\theta)}{\tan^2(\theta)} \sec^2(\theta) \, d\theta \).

Step 3: Simplify the Integral in Terms of \( \theta \)

Rewrite the integral as \( \int \frac{\sec^3(\theta)}{\tan^2(\theta)} \, d\theta \). Note that \( \tan^2(\theta) = \sec^2(\theta) - 1 \), so \( \frac{1}{\tan^2(\theta)} = \frac{1}{sec^2(\theta) - 1} \). The integral becomes complex for direct integration, seeking a change in variable expressions.

Step 4: Re-Integration by Another Substitution

Let \( u = \tan(\theta) \), then \( du = \sec^2(\theta) \, d\theta \). The expression becomes \( \int \frac{\sec(\theta)}{u^2} \sec^2(\theta) \frac{du}{\sec^2(\theta)} = \int \frac{sec(\theta)}{u^2} \, du \). Simplify to \( \int \frac{1}{u^2} \, du \).

Step 5: Compute the Simplified Integral

The integral \( \int \frac{1}{u^2} \, du \) is straightforward. It evaluates to \( -\frac{1}{u} + C \), where \( C \) is the integration constant.

Step 6: Substitute Back to Original Variable

Recall \( u = \tan(\theta) \) and \( x = \tan(\theta) \). Therefore, \( \frac{1}{u} = \frac{1}{\tan(\theta)} = \cot(\theta) = \frac{x}{\sqrt{1+x^2}} \). Substitute back to express the solution in terms of \( x \): The integrated result is \( -\frac{x}{\sqrt{1+x^2}} + C \).

Key Concepts

Integration Techniques

Integration techniques are special methods used to solve complex integrals, making them more manageable. One powerful technique is **trigonometric substitution**, which is very helpful when you have an integral involving a square root and polynomials, as seen in our exercise.For example, trigonometric substitution leverages trigonometric identities and functions to transform the integral into a simpler form. In the case of the integral \( \int \frac{\sqrt{1+x^{2}}}{x^{2}} \, dx \), using the substitution \( x = \tan(\theta) \) is effective. This is because the identity \( 1 + \tan^2(\theta) = \sec^2(\theta) \) simplifies the expression under the square root.Here's a quick summary of when to use different integration techniques:

• **Substitution**: Useful when you can transform the integral into an easily integrable form by changing variables.
• **Trigonometric substitution**: Best for integrals involving expressions like \( \sqrt{a^2 + x^2} \) or \( \sqrt{x^2 - a^2} \).
• **Integration by parts**: Applicable when the product of two functions is integrated.

Integration can sometimes feel tricky, but using the right technique will make it straightforward.

Calculus

Calculus is a branch of mathematics that studies continuous change and can be broadly divided into **differential calculus** and **integral calculus**. In our context, integral calculus is of utmost importance as it deals with finding the total accumulation of quantities, which is represented by integrals. The exercise we're looking at requires the use of some core concepts of calculus. It demands not only knowledge of integration but also understanding of derivatives because integration is the inverse operation of differentiation. Here's a brief look into why calculus is essential:

• **Understanding motion**: Calculus helps in finding exact values of distance and velocity.
• **Optimizing functions**: Allows finding maximum or minimum values of functions which is crucial in many real-world applications.
• **Solving equations**: Enables solutions to complex differential equations, which describe numerous phenomena in engineering and physics.

In our task, calculus provides the broad framework within which integration techniques work to solve definite and indefinite integrals.

Definite Integrals

Definite integrals are used to calculate the **net area** under a curve from one point to another along the x-axis. It's not just about finding an antiderivative but also about evaluating it over a specific interval.In contrast to **indefinite integrals** that yield a family of functions (along with an integration constant \( C \)), definite integrals provide a numerical value as a result.In our specific problem, while we demonstrated the process of solving the integral, note that definite integrals have distinct properties:

• **Boundaries**: Definite integrals are evaluated over a specific range \([a, b]\).
• **Application**: They are fundamental in calculating quantities like areas, probabilities, and even physical properties like mass and charge distributions.

To evaluate a definite integral, after performing integration, substitute the upper and lower limits into the integrated function to find the net result. Remember, definite integrals are powerful tools in both pure and applied mathematics, providing essential solutions in many fields.