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Magazine Chapter 28: Problem 41
Solve the given problems by integration. Show that \(\int \sin x \cos x d x\) can be integrated in two ways. Explain the difference in the answers.
Short Answer
Expert verified
The integral \( \int \sin x \cos x \, dx \) can be solved as \( \frac{1}{2} \sin^2 x + C \) or \(-\frac{1}{4} \cos 2x + C \), differing only by a constant.
Step by step solution
01
Recognize the Integration By Substitution
First, we can use the substitution method to integrate \( \int \sin x \cos x \, dx \). Identify \( u = \sin x \), then \( du = \cos x \, dx \). This substitution makes the integral \( \int u \, du \).
02
Perform the Integration for Substitution
The integral \( \int u \, du \) becomes \( \frac{u^2}{2} + C \). Re-substitute \( u = \sin x \) back into the equation to get the integral result: \[ \frac{1}{2} \sin^2 x + C \].
03
Recognize the Use of Trigonometric Identity
Another method involves using the double angle formula: \( \sin 2x = 2 \sin x \cos x \). Solve for \( \sin x \cos x \) to get \( \frac{1}{2} \sin 2x \). Substitute this into the original integral: \( \int \sin x \cos x \, dx = \frac{1}{2} \int \sin 2x \, dx \).
04
Integrate Using the Double Angle Formula
Carry out the integration of \( \frac{1}{2} \int \sin 2x \, dx \), which results in \(-\frac{1}{4} \cos 2x + C\). This integral represents the same function as before but in a different form.
05
Compare Two Methods
The differences in expressions arise because the integral method using substitution gives \( \frac{1}{2} \sin^2 x + C \) and using the double angle formula produces \(-\frac{1}{4} \cos 2x + C \). They differ formally but represent the same family of functions, differing only by a constant.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Integration by Substitution
Integration by substitution, also known as the u-substitution method, is a technique used to simplify the process of finding integrals. The key idea is to substitute a part of the integrand (the function being integrated) with a single variable, usually denoted as "u". This transforms the integral into a simpler form.
- Identify a function within the integrand that you can substitute with "u". In this case, we choose \( u = \sin x \).
- Differentiate \( u \) with respect to \( x \), which gives \( du = \cos x \, dx \).
- Replace the original integrand with \( u \) to change the integral to \( \int u \, du \).
Trigonometric Integrals
Trigonometric integrals involve functions like sine, cosine, and tangent. These can often be tricky to integrate directly. A solid understanding of trigonometric identities can simplify the process significantly.
In this problem, the integral \( \int \sin x \cos x \, dx \) can be approached using trigonometric identities. Specifically, the double angle identity \( \sin 2x = 2 \sin x \cos x \) allows for a straightforward modification:
In this problem, the integral \( \int \sin x \cos x \, dx \) can be approached using trigonometric identities. Specifically, the double angle identity \( \sin 2x = 2 \sin x \cos x \) allows for a straightforward modification:
- Rearrange to express \( \sin x \cos x \) as \( \frac{1}{2} \sin 2x \).
- Substitute into the integral to transform it: \( \int \sin x \cos x \, dx = \frac{1}{2} \int \sin 2x \, dx \).
- This creates an integral that is often simpler to handle.
Double Angle Formula
The double angle formula is a specific type of trigonometric identity that expresses trigonometric functions of double angles in terms of square or multiple trigonometric functions of single angles.
For instance, the formula \( \sin 2x = 2 \sin x \cos x \) is crucial in transforming the product of sine and cosine into a single trigonometric function. Here's how it helps:
For instance, the formula \( \sin 2x = 2 \sin x \cos x \) is crucial in transforming the product of sine and cosine into a single trigonometric function. Here's how it helps:
- The original integral \( \int \sin x \cos x \, dx \) can be transformed using the identity \( \sin 2x = 2 \sin x \cos x \).
- This gives \( \sin x \cos x = \frac{1}{2} \sin 2x \), simplifying the integral to \( \frac{1}{2} \int \sin 2x \, dx \).
- Perform the integration to find \(-\frac{1}{4} \cos 2x + C \).
Constants of Integration
When performing indefinite integrals, the constant of integration \( C \) is an essential component. It represents an arbitrary constant that accounts for all possible vertical shifts of the antiderivative function.
- The expression \( \int f(x) \, dx = F(x) + C \) indicates that any constant could have been part of the original derivative function.
- In this problem, both methods for solving \( \int \sin x \cos x \, dx \) result in expressions: \( \frac{1}{2} \sin^2 x + C \) and \(-\frac{1}{4} \cos 2x + C \).
- The difference in the constants \( C \) reflects that the solutions differ only by a constant value.
