Problem 61 - Thomas Calculus

Question

It looks as if we can integrate 2 sin \(x \cos x\) with respect to \(x\) in three different ways: $$ \begin{aligned} \text { a. } \int 2 \sin x \cos x d x &=\int 2 u d u \quad u=\sin x \\ &=u^{2}+C_{1}=\sin ^{2} x+C_{1} \\ \text { b. } \int 2 \sin x \cos x d x &=\int-2 u d u \quad u=\cos x \\ &=-u^{2}+C_{2}=-\cos ^{2} x+C_{2} \\\ \text { c. } \int 2 \sin x \cos x d x &=\int \sin 2 x d x \quad 2 \sin x \cos x=\sin 2 x \\ &=-\frac{\cos 2 x}{2}+C_{3} \end{aligned} $$ Can all three integrations be correct? Give reasons for your answer.

Step-by-Step Solution

Answer

Yes, all three integrations are correct, differing only by a constant.\n

Step 1: Review the Problem

Identify the function to be integrated: \( ext{a. } ext{b. } ext{c. } \ ext{and } \int 2 \sin x \cos x \, dx \). Note that you have three different ways of rewriting and integrating this same integral using different substitutions or identities.

Step 2: Use the Substitution Method for Option A

For the first method, \( u = \sin x \). Therefore, \( du = \cos x \, dx \). Substitute to get \( \int 2 \sin x \cos x \, dx = \int 2u \, du \). Integrating gives \( u^2 = \sin^2 x \), leading to the integral \( \sin^2 x + C_1 \).

Step 3: Apply Substitution for Option B

In the second method, set \( u = \cos x \), resulting in \( du = -\sin x \, dx \). Substitute to derive \( \int 2 \sin x \cos x \, dx = \int -2u \, du \). Integrating yields \( -u^2 = -\cos^2 x \), which results in \( -\cos^2 x + C_2 \).

Step 4: Use the Double Angle Method for Option C

Observe that \( 2 \sin x \cos x = \sin 2x \). Substitute this identity into the integral; it becomes \( \int \sin 2x \, dx \). Integrating gives \( -\frac{\cos 2x}{2} + C_3 \).

Step 5: Compare the Results

Recall the trigonometric identity \( \sin^2 x + \cos^2 x = 1 \). This means that \( \sin^2 x = 1 - \cos^2 x \), providing consistency between the results of (a) and (b). Result from (c), \( -\frac{\cos 2x}{2} \), can be rewritten using \( \cos(2x) = 1 - 2\sin^2 x \) or \( \cos(2x) = 2\cos^2 x - 1 \). Transform these to match either option (a) or (b). All three results are correct, only constants of integration differ, representing the same family of functions by different constants.

Key Concepts

Trigonometric Identities

Trigonometric identities are mathematical expressions that relate the angles and sides of a right triangle in different ways. These identities are extremely useful when integrating trigonometric functions as they allow us to simplify complex expressions into more manageable forms. For example, in the problem provided, we use the double angle identity: \( 2 \sin x \cos x = \sin 2x \). This identity transforms the original integral into a simpler form that can be easily solved.
Here are a few key trigonometric identities that come in handy when performing integration:

• \( \sin^2 x + \cos^2 x = 1 \): The Pythagorean Identity, useful for replacing one function in terms of the other.
• Double Angle Identities: These involve functions like sine, cosine, and tangent to express angles twice as large.
• Other transformations include \( \tan x = \frac{\sin x}{\cos x} \) and \( \cot x = \frac{\cos x}{\sin x} \).

Knowing these identities allows you to convert complex trigonometric expressions into simpler forms that are easier to integrate. Always try to identify the simplest form when dealing with trigonometric integrals.

Substitution Method

The substitution method, also known as \(u\)-substitution, is a powerful technique used to simplify integrals. When a function is composed of another function and its derivative, substitution can transform the integral into an easier form.
In the given exercise, different substitutions were used to solve the integral \( \int 2 \sin x \cos x \, dx \):

• For option (a), the substitution \( u = \sin x \) simplifies the problem as \( du = \cos x \, dx \), leading to \( \int 2u \, du \).
• In option (b), the substitution \( u = \cos x \) results in \( du = -\sin x \, dx \), transforming the integral into \( \int -2u \, du \).

When using substitution, the goal is to transform the original integral into a simpler one by making a suitable substitution. After finding the integral of the simpler function, don't forget to substitute back in terms of the original variable.
The key advantage of this method is that it potentially changes a complicated integral into a manageable one. When faced with an integral, always consider if the substitution method could simplify it.

Definite and Indefinite Integrals

Integration, in calculus, is a fundamental concept that comes in two main forms: definite and indefinite integrals.

• Indefinite Integrals: These represent a family of functions and include a constant of integration \( C \). In the exercise provided, examples of indefinite integrals are given in options (a), (b), and (c). Indefinite integrals give you a general solution indicating all possible antiderivatives of a function.
• Definite Integrals: These calculate the net area under a curve between two bounds, providing a number as a result. Though definite integrals aren't the focus of this problem, they are just as important in understanding the full scope of integration.

The main distinction is that indefinite integrals concern general solutions with a family of curves, whereas definite integrals provide specific values over an interval. In any exercise involving integration, knowing whether you're dealing with definite or indefinite integrals will guide your solution approach.
To conclude, integrating functions, whether definite or indefinite, serves the purpose of determining accumulations, areas, or general solutions to differential equations. Understanding both allows for more effective problem-solving in calculus.