Problem 71 - Calculus Single Variable

Question

By making the substitution \(u=\sin (x),\) show that $$\int \sin (x) \cos (x) d x=\frac{1}{2} \sin ^{2}(x)+C$$ By making the substitution \(u=\cos (x),\) show that $$\int \sin (x) \cos (x) d x=-\frac{1}{2} \cos ^{2}(x)+C $$By using the formula for \(\sin (2 x)\), show that $$\int \sin (x) \cos (x) d x=-\frac{1}{4} \cos (2 x)+C$$ Explain why these answers are not contradictory.

Step-by-Step Solution

Answer

These results show equivalent expressions for the integral, differing only by constants due to substitution choice.

Step 1: Substitution with u = sin(x)

Begin by letting \(u = \sin(x)\). Then the derivative \(du = \cos(x)\,dx\). Substituting into the integral, we have:\[ \int \sin(x) \cos(x)\, dx = \int u\, du \]The integral of \(u\) is \(\frac{1}{2}u^2 + C\). Substituting back \(u = \sin(x)\), the solution is:\[ \frac{1}{2}\sin^2(x) + C \]

Step 2: Substitution with u = cos(x)

Here, let \(u = \cos(x)\). Then \(du = -\sin(x)\,dx\) or \(-du = \sin(x)\,dx\). Substituting into the integral, we get:\[ \int \sin(x) \cos(x)\, dx = \int -u\, du \]This integral evaluates to \(-\frac{1}{2}u^2 + C\). Substituting back \(u = \cos(x)\), we have:\[ -\frac{1}{2}\cos^2(x) + C \]

Step 3: Using the formula for sin(2x)

Recall the identity \( \sin(2x) = 2\sin(x)\cos(x) \). Therefore, \( \sin(x)\cos(x) = \frac{1}{2}\sin(2x) \).Substituting into the integral, we have:\[ \int \sin(x)\cos(x)\, dx = \int \frac{1}{2} \sin(2x)\, dx \]Let \(v = 2x\), so \(dv = 2\,dx\) or \(dx = \frac{1}{2}\,dv\). Therefore, the integral becomes:\[ \frac{1}{2} \int \sin(2x)\, dx = -\frac{1}{4}\cos(2x) + C \]

Step 4: Analyzing the Answers

All expressions represent the same family of antiderivatives resulting from different perspectives of integration, differing only in their constant of integration. Each form can be shown to differ from the others by a constant using trigonometric identities, as integration is not unique due to the arbitrary constant \(C\). This shows how results that look different can be equivalent through trigonometric relationships and constant adjustments.

Key Concepts

Integration by Substitution

Integration by substitution is a powerful tool that simplifies complex integrals by changing variables. This method transforms an integral into a new form that's often easier to evaluate.
Here's how it works:1. **Choose a Substitution:** Identify a part of the integral to replace with a new variable. For example, if you substitute using the trig function, like \(u = \sin(x)\), you're looking for parts of the integral related to \(\sin(x)\).2. **Differentiate to find \(du\):** Take the derivative of the new variable in terms of \(x\). For \(u = \sin(x)\), this gives \(du = \cos(x)\ dx\).3. **Substitute:** Replace all instances of the part you substituted with your new variable. Also, replace the differential \(dx\) using \(du\).
In the example: - Substituting \(u = \sin(x)\), changes \(x\) into \(du\). The integral becomes \(\int u\ du\) leading to \(\frac{1}{2}u^2 + C\).- Replacing back \(u\) with the original terms gives \(\frac{1}{2}\sin^2(x) + C\).
This substitution method streamlines the process by transforming complex expressions into manageable ones.

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that hold true for all values of the involved variables. These identities are crucial for simplifying expressions and solving integrals involving trigonometric functions.
Some vital identities:- **Pythagorean Identity:** \(\sin^2(x) + \cos^2(x) = 1\)- **Double Angle Identity for Sine:** \(\sin(2x) = 2\sin(x)\cos(x)\) These identities allow us to express trigonometric integrals in different ways. For example:- Using \(\sin(2x) = 2\sin(x)\cos(x)\), the integral \(\int \sin(x)\cos(x)\ dx\) can be re-written as \(\frac{1}{2}\int \sin(2x)\ dx\).
These transformations enable easier integration by reducing integrals to simpler or different forms, solving them more easily.
Understanding and applying these identities allow for multiple methods of solving integrals, though they might appear different initially. However, due to trigonometric relationships and constants, they can be shown to be equivalent.

Integral Calculus

Integral calculus is a branch of calculus concerned with the concept of integration, used to determine the area under curves, among other uses. There are indefinite and definite integrals:- **Indefinite Integrals:** Are without limits and include a constant of integration \(C\) because the antiderivative isn't unique.- **Definite Integrals:** Are evaluated over an interval and result in a numerical value representing the area under a curve between two points.
In integral calculus:- The **process of finding the integral** is known as anti-differentiation or finding the antiderivative.- **Integration techniques** like substitution or using identities help simplify complicated integrals.
For instance, in the exercise example:- **Indefinite integrals** were solved using substitution and trigonometric identities, showcasing how each method leads to an antiderivative of the same original function.- **Different methods can lead to forms that look different**; however, these differences are reconciled through understanding constant integration and applicable trigonometric relationships, demonstrating how integration results can vary yet remain valid across methods. Thus, integral calculus provides the groundwork for solving complex problems by breaking them down into simpler, manageable operations.