Problem 24 - Thomas Calculus

Question

Is either of the following equations correct? Give reasons for your answers. a. \(\frac{1}{\cos x} \int \cos x d x=\tan x+C\) b. \(\frac{1}{\cos x} \int \cos x d x=\tan x+\frac{C}{\cos x}\)

Step-by-Step Solution

Answer

Equation (b) is correct; equation (a) is not correct for non-zero \(C\).

Step 1: Evaluate the Integral

First, we evaluate the integral \(\int \cos x \, dx\). The integral of \(\cos x\) with respect to \(x\) is \(\sin x\). Thus, the integral is \(\sin x + C\) where \(C\) is the constant of integration.

Step 2: Simplify Equation a

Now we simplify the equation in part (a), \(\frac{1}{\cos x} \int \cos x \, dx = \tan x + C\).Substituting the result from Step 1, this becomes \(\frac{1}{\cos x}(\sin x + C)\). Simplifying gives:\[\frac{\sin x}{\cos x} + \frac{C}{\cos x} = \tan x + \frac{C}{\cos x}.\]This matches \(\tan x + C\) only if \(C = 0\), not generally for any \(C\). Thus, equation (a) is not correct for \(C eq 0\).

Step 3: Simplify Equation b

Next, we verify equation (b), \(\frac{1}{\cos x} \int \cos x \, dx = \tan x + \frac{C}{\cos x}\).Substituting the result from Step 1, this becomes \(\frac{1}{\cos x}(\sin x + C)\), which simplifies to:\[\frac{\sin x}{\cos x} + \frac{C}{\cos x} = \tan x + \frac{C}{\cos x}.\]Equation (b) matches this simplification exactly, hence it is correct.

Key Concepts

Calculus

Calculus is a vast area of mathematics that deals with continuous change. It primarily revolves around two fundamental concepts: differentiation and integration. While differentiation focuses on finding the rate of change, integration is about finding the total accumulation of quantities. Integrals can be thought of as the reverse process of derivatives. When working with integrals, we often seek to find the area under a curve. In calculus, solving integrals is a crucial skill that allows us to handle functions in more complex situations.

Integrals come in different types, but in this particular problem, we're dealing with definite and indefinite integrals. An indefinite integral is expressed without upper and lower limits and typically includes a constant of integration, whereas definite integrals have limits and provide a numerical value as the result. Recognizing the correct methods and rules to simplify these integrals accurately is essential for solving calculus problems.

Trigonometric Integrals

Trigonometric integrals involve integrating expressions with trigonometric functions like sine, cosine, tangent, and others. These types of integrals can pose a challenge due to their oscillatory nature, but understanding the basic antiderivatives of trigonometric functions goes a long way in solving them.

For example, when you seek to integrate \(\sin x\), you get \(\-\cos x + C\). Similarly, integrating \(\cos x\) results in \(\sin x + C\). Recognizing these relationships gives you the tools to tackle more complex integrals involving trigonometric functions. In the case of this exercise, the integral \(\int \cos x \, dx\) simplifies directly to \(\sin x + C\). The next step is using algebraic techniques to further manipulate these expressions as shown in steps 2 and 3, connecting trigonometric identities with calculus concepts.

Constant of Integration

The constant of integration, often denoted by \(C\), is an unavoidable component when evaluating indefinite integrals. This constant represents an unknown constant amount added to the function, arising because indefinite integrals are families of functions that can be shifted vertically on a graph. Every antiderivative of a function is perfect only up to an additive constant, hence the inclusion of \(C\).

In the solution provided, variations of \(C\) affect the final expression, and only specific conditions make equation (a) satisfy the given integral expression. By setting \(C = 0\), equation (a) can be coherent, but this doesn't hold in general cases where \(C\) is arbitrary. The constant of integration plays a significant role in defining the set of solutions to an integral problem, reminding us of the diversity in antiderivative solutions.