Problem 13 - Calculus of a Single Variable

Question

Finding an Indefinite Integral In Exercises \(1-20\) , find the indefinite integral. $$ \int \frac{\sec ^{2} x}{\sqrt{25-\tan ^{2} x}} d x $$

Step-by-Step Solution

Answer

The indefinite integral of the initial function is \(\tan x + C\).

Step 1: Simplify the Integral

Notice that the denominator contains the expression \(25 - \tan^2x\), which can be replaced by \(\sec^2x - \tan^2x\), based on the trigonometric identity \(1 + \tan^2x = \sec^2x\). The integral becomes \(\int \frac{\sec^2x}{\sqrt{\sec^2x - \tan^2x}} dx\).

Step 2: Substitution Method

We can solve the simplified integral using the substitution method. Let \(u = \tan x\). Then \(du = \sec^2xdx\), and \(dx = \frac{du}{\sec^2x}\). The integral becomes \(\int \frac{du}{\sqrt{\sec^2x - u^2}}\). Again, using the Pythagorean identity \(1 + \tan^2x = \sec^2x\), we get \(sec^2x = 1 + u^2\) so the integral now becomes \(\int \frac{1}{\sqrt{{1 + u^2} - u^2}} du = \int \frac{1}{\sqrt{1}} du\).

Step 3: Calculate the Indefinite Integral

The integral \(\int du\) is just \(u + C\). Substitute back \(u = \tan x\), the final answer is \(\tan x + C\).

Key Concepts

Trigonometric Identities

Trigonometric identities are vital tools in mathematics that allow us to simplify and manipulate expressions involving trigonometric functions. They are equations that are true for all values of the variable where both sides of the equation are defined. Some common trigonometric identities include:

• The Pythagorean identity: \(\sin^2 x + \cos^2 x = 1\)
• The tangent and secant identity: \(1 + \tan^2 x = \sec^2 x\)
• The cotangent and cosecant identity: \(1 + \cot^2 x = \csc^2 x\)

In the given exercise, the identity \(1 + \tan^2 x = \sec^2 x\) is used to simplify the expression in the denominator. This particular identity helps us transform complex trigonometric expressions into simpler forms that are more manageable and easier to integrate.

Substitution Method

The substitution method in calculus is a powerful technique used to evaluate integrals. It involves changing the variable in the integral to simplify the expression, similar to using a change of variables. This method is particularly useful when dealing with trigonometric integrals or integrals involving composite functions.When using the substitution method, follow these steps:

• Identify a substitution that can simplify the integral. This often involves choosing a 'u' that makes the differential 'du' appear elsewhere in the integral.
• Write down the relationship between 'dx' and 'du'.
• Substitute 'u' in the integral, simplifying the expression if possible.
• After integrating, substitute back the original variable.

In the exercise, the substitution \(u = \tan x\) simplifies the integral because \(du = \sec^2 x dx\), turning a complex trigonometric integral into a much simpler form.

Pythagorean Identity

The Pythagorean identity is one of the foundational identities in trigonometry. It expresses the fundamental relationship between sine, cosine, and other trigonometric functions. The most well-known Pythagorean identity is \(\sin^2 x + \cos^2 x = 1\). From this, we can derive other useful identities:

• \(1 + \tan^2 x = \sec^2 x\)
• \(1 + \cot^2 x = \csc^2 x\)

In the exercise, the Pythagorean identity \(1 + \tan^2 x = \sec^2 x\) is rearranged to express \(\sec^2 x\) in terms of \(\tan^2 x\). This allows the simplification of the expression under the square root, ultimately aiding in the integration process. Recognizing and applying these identities correctly is crucial in solving integrals efficiently.

Integral Calculus

Integral calculus is a significant part of calculus that deals with integrals and their properties. It focuses on finding antiderivatives of functions, a process called integration, which is the inverse operation of differentiation. There are two main types of integrals:

• Indefinite Integrals: These represent families of functions and include a constant of integration, denoted by \(C\).
• Definite Integrals: These calculate the area under a curve within a specified interval and yield a numerical value.

Indefinite integrals, like those in the exercise, do not have specific limits and give a general form of the antiderivative. The solution of the integral \(\int du\) yields \(u + C\), where \(C\) is the constant of integration, reminding us of the family of curves that represent the antiderivative. The final result, \(\tan x + C\), shows that the antiderivative of the given expression is related to the tangent function, underlining the profound connections between trigonometric functions and their integrals.