Problem 49 - Thomas Calculus

Question

Evaluate the integrals in Exercises \(47-70\) $$ \int \frac{d x}{17+x^{2}} $$

Step-by-Step Solution

Answer

The integral is \( \frac{1}{\sqrt{17}} \tan^{-1}\left(\frac{x}{\sqrt{17}}\right) + C \).

Step 1: Recognize the Integral Form

The given integral is \( \int \frac{dx}{17+x^2} \). This integral follows the standard form \( \int \frac{dx}{a^2 + x^2} = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C \), where \( a^2 = 17 \).

Step 2: Identify the Constant 'a'

In this integral, we need to find \( a \) such that \( a^2 = 17 \). Solving this gives \( a = \sqrt{17} \).

Step 3: Substitute and Apply the Formula

Substitute \( a = \sqrt{17} \) into the standard integral formula. This gives us: \[ \int \frac{dx}{17+x^2} = \frac{1}{\sqrt{17}} \tan^{-1}\left(\frac{x}{\sqrt{17}}\right) + C \]

Step 4: Write the Final Answer

The integral of the function \( \frac{1}{17+x^2} \) with respect to \( x \) is \( \frac{1}{\sqrt{17}} \tan^{-1}\left(\frac{x}{\sqrt{17}}\right) + C \), where \( C \) is the constant of integration.

Key Concepts

Definite Integrals

Definite integrals are the key to finding the area under a curve between two specific points on the x-axis. When you have boundaries for integration, like from point \(a\) to \(b\), you are dealing with a definite integral. The notation for this is \( \int_{a}^{b} f(x) \, dx \). This type of integral not only calculates the area but also accounts for the direction of the area; if the function dips below the x-axis, that area is considered negative.

Here’s why they are important:

• Provide a way to calculate accumulated quantities, like distance, area, or volume.
• Useful in physics and engineering for finding things like work done by a force.

When evaluating a definite integral, you find the antiderivative (or indefinite integral) first, and then you evaluate it at the upper and lower limits: \( F(b) - F(a) \). This gives you the net area between the function and the x-axis over the specified interval.

Indefinite Integrals

Indefinite integrals are all about finding functions whose derivatives produce the given function. It's what you do when you're told “find the antiderivative” of a function. So, when you see \(\int f(x) \, dx\), you are looking for a function \(F(x)\) such that \(F'(x) = f(x)\).

Some key points:

• Indefinite integrals include a constant of integration \(C\) because derivatives of constant functions are zero. So, any constant can be added to the antiderivative without affecting the derivative result.
• They are a fundamental tool in solving differential equations.

For example, if you are given \(\int \frac{1}{17+x^2} \, dx\), the standard form \(\int \frac{dx}{a^2 + x^2} = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C\) is essential for finding the antiderivative. This allows us to solve problems involving inverse trigonometric functions as seen in the step-by-step solution.

Inverse Trigonometric Functions

Inverse trigonometric functions are used when you need to find angles rather than direct trigonometric values. Common inverse trig functions include \(\arcsin\), \(\arccos\), and \(\arctan\). These functions are particularly useful in calculus when working with integrals involving expressions like \(a^2 + x^2\).

Why are they important?

• The antiderivative of certain rational functions can be expressed in terms of these inverse functions.
• They help in expressing the angles in trigonometric equations, which is often necessary in application problems.

For instance, in the integral \(\int \frac{dx}{17+x^2}\), recognizing the inverse tangent form is crucial. The solution \(\frac{1}{\sqrt{17}} \tan^{-1}\left(\frac{x}{\sqrt{17}}\right) + C\) involves the inverse tangent, which reveals how integral calculus interacts with trigonometric and inverse functions to solve problems.