Problem 66 - Applied Calculus

Question

(a) Find \(\int \sin \theta \cos \theta d \theta\) (b) You probably solved part (a) by making the substitution \(w=\sin \theta\) or \(w=\cos \theta .\) (If not, go back and do it that way.) Now find \(\int \sin \theta \cos \theta d \theta\) by making the other substitution. (c) There is yet another way of finding this integral which involves the trigonometric identities $$\begin{aligned}\sin (2 \theta) &=2 \sin \theta \cos \theta \\\\\cos (2 \theta) &=\cos ^{2} \theta-\sin ^{2} \theta\end{aligned}$$ Find \(\int \sin \theta \cos \theta d \theta\) using one of these identities and then the substitution \(w=2 \theta\) (d) You should now have three different expressions for the indefinite integral \(\int \sin \theta \cos \theta d \theta .\) Are they really different? Are they all correct? Explain.

Step-by-Step Solution

Answer

The expressions for the integral in all three parts are equivalent up to a constant.

Step 1: Solution to Part (a)

First, let's use the substitution \( w = \sin \theta \). Then, the derivative \( dw = \cos \theta \, d\theta \). Thus, the integral becomes:\[\int \sin \theta \cos \theta \, d\theta = \int w \, dw\]Integrating with respect to \( w \), we get:\[\frac{w^2}{2} + C = \frac{\sin^2 \theta}{2} + C\]

Step 2: Solution to Part (b)

Now, let's use the substitution \( w = \cos \theta \). The derivative gives \( dw = -\sin \theta \, d\theta \), or \( -dw = \sin \theta \, d\theta \). Thus, the integral transforms to:\[\int \sin \theta \cos \theta \, d\theta = \int -w \, dw\]Integrating with respect to \( w \), result in:\[-\frac{w^2}{2} + C = -\frac{\cos^2 \theta}{2} + C\]

Step 3: Solution to Part (c) Using Trigonometric Identity

Using the identity \( \sin(2\theta) = 2 \sin \theta \cos \theta \), we can rewrite the original integral:\[\int \sin \theta \cos \theta \, d\theta = \frac{1}{2} \int \sin(2\theta) \, d\theta\]Now, let \( w = 2\theta \), which implies \( dw = 2 \, d\theta \) or \( d\theta = \frac{dw}{2} \). The integral becomes:\[\frac{1}{2} \int \sin(w) \frac{dw}{2} = \frac{1}{4} \int \sin(w) \, dw\]Integrating, we get:\[-\frac{1}{4} \cos(w) + C = -\frac{1}{4} \cos(2\theta) + C\]

Step 4: Compare Results to Part (d)

From steps 1, 2, and 3 we have the results:1. \( \int \sin \theta \cos \theta \, d\theta = \frac{\sin^2 \theta}{2} + C \)2. \( \int \sin \theta \cos \theta \, d\theta = -\frac{\cos^2 \theta}{2} + C \)3. \( \int \sin \theta \cos \theta \, d\theta = -\frac{1}{4} \cos(2\theta) + C \)To check if these are equivalent, use the identities \(\sin^2 \theta = 1 - \cos^2 \theta\) and \(\cos(2\theta) = \cos^2 \theta - \sin^2 \theta\). Each expression simplifies to comparable forms involving different constants, indicating they represent the same family of functions up to constant differences.

Key Concepts

Integration Techniques

Integration techniques are methods used to solve integrals, which are essentially the reverse process of differentiation. These techniques help us find the antiderivative of a function. There are several methods to tackle integrals:

• Direct Integration: Used when the integral of the function is straightforward and can be found directly.
• Substitution Method: A technique where we substitute a part of the integral to simplify the integration process.
• Integration by Parts: Used when the original function is a product of two functions; it applies the formula derived from the product rule of differentiation.
• Trigonometric Substitution: Used specifically when dealing with integrals involving radical expressions.

For the exercise, we are using substitution methods and a trigonometric identity to solve the integral \( \int \sin \theta \cos \theta \, d \theta \). Instead of directly integrating, these techniques simplify the expression to make the integration process more manageable.

Substitution Method

The substitution method is a critical tool in solving integrals, especially when direct integration isn't straightforward. It involves substituting a part of the integrand with a new variable, simplifying the expression, performing the integration, and then substituting back the original variable.
In the given exercise, we use substitution in different ways:

• First, by letting \( w = \sin \theta \), where \( dw = \cos \theta \, d\theta \). The original integral \( \int \sin \theta \cos \theta \, d\theta \) becomes \( \int w \, dw \), which is easier to solve.
• Second, using \( w = \cos \theta \), we find \( dw = -\sin \theta \, d\theta \). The integral then changes to \( \int -w \, dw \).

Both substitutions turn the trigonometric integral into a simple polynomial integral, making it easier to find the antiderivative. Once the integration is performed, we must remember to replace the substituted variable back to the original function term.

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that hold true for every value of the variable where both sides of the identity are defined. They are particularly useful in simplifying integrals and solving trigonometric equations.
In this exercise, the identity \( \sin(2\theta) = 2 \sin \theta \cos \theta \) was used. By recognizing this identity, the integral \( \int \sin \theta \cos \theta \, d\theta \) could be rewritten in a more manageable form using \( \sin(2\theta) \).
To solve the problem:

• The expression \( 2 \sin \theta \cos \theta \) was identified as \( \sin(2\theta) \), leading to \( \frac{1}{2} \int \sin(2\theta) \, d\theta \).
• By substituting \( w = 2\theta \) and \( dw = 2 \, d\theta \), the integral becomes \( \frac{1}{4} \int \sin(w) \, dw \).

Trigonometric identities not only simplify the integral but also allow the use of double angle formulas or other simplifications.