Problem 269 - Calculus Volume 1

Question

In the following exercises, find the antiderivative using the indicated substitution. $$ \int \cos ^{3} \theta d \theta ; u=\sin \theta\left(\operatorname{Hint} \cos ^{2} \theta=1-\sin ^{2} \theta\right) $$

Step-by-Step Solution

Answer

The antiderivative is \( \sin \theta - \frac{(\sin \theta)^3}{3} + C \).

Step 1: Substitute the Trigonometric Identity

First, use the hint provided in the problem to express \( \cos^2 \theta \) in terms of \( \sin \theta \). We have \( \cos^2 \theta = 1 - \sin^2 \theta \). Therefore, \( \cos^3 \theta = \cos \theta \cdot \cos^2 \theta = \cos \theta (1 - \sin^2 \theta) \).

Step 2: Express in Terms of u

Since \( u = \sin \theta \), we express the integral in terms of \( u \). Therefore, \( \cos \theta = \sqrt{1-u^2} \), but note that with the substitution \( du = \cos \theta \, d\theta \), we can directly replace \( \cos \theta \, d\theta = du \). The expression becomes \[ \int \cos^3 \theta \, d\theta = \int \cos \theta (1 - \sin^2 \theta) \, d\theta = \int \cos \theta (1 - u^2) \, d\theta. \] But, recognizing \( \cos \theta \, d\theta = du\), the integral becomes \[ \int (1-u^2) \, du. \]

Step 3: Integrate with Respect to u

Now that we have the integral in terms of \( u \), integrate with respect to \( u \): \[ \int (1 - u^2) \, du = \int 1 \, du - \int u^2 \, du. \] This yields \[ u - \frac{u^3}{3} + C, \] where \( C \) is the constant of integration.

Step 4: Substitute Back to Original Variable

Finally, substitute back \( u = \sin \theta \) into the antiderivative: \[ u - \frac{u^3}{3} + C = \sin \theta - \frac{(\sin \theta)^3}{3} + C. \] This gives the antiderivative in terms of \( \theta \).

Key Concepts

Trigonometric Substitution

Trigonometric substitution is a clever technique used in calculus to simplify integrals involving trigonometric functions. In our exercise, we're dealing with an integral that involves powers of cosine. The hint suggests using a trigonometric identity to transform part of the expression. This identity helps us rewrite

• \( \cos^2 \theta = 1 - \sin^2 \theta \)
• \( \cos^3 \theta = \cos \theta (1 - \sin^2 \theta) \)

By substituting \( \cos^2 \theta \) with \( 1 - \sin^2 \theta \), the expression becomes much simpler. This sets the stage for changing variables, a key part of trigonometric substitution.To perform the substitution, we set \( u = \sin \theta \). This choice of substitution leverages the relationship between sine and cosine. Trigonometric substitution is particularly useful for integrals that turn into functions of sine and cosine, as it simplifies the variable transformation process.

Integration Techniques

Integration techniques include various methods used to solve integrals that can't be immediately evaluated. In the given problem, we've used a combination of substitutions and algebraic manipulation to simplify the process.

Here's how it works:

• First, we use trigonometric identities to rewrite the integral in a simpler form.
• We then perform a substitution so that the integral can be expressed in terms of a single variable \( u \), easing the computation.

Once the expression is in terms of \( u \), our integration task becomes a breeze. The integral reduces to solving \[ \int (1-u^2) \, du \]
Integration techniques like substitution are invaluable because they turn complex-looking integrals into simpler tasks. They often produce expressions that are algebraically manageable, allowing us to neatly find the antiderivatives.

Calculus Exercises

Calculus exercises often involve multiple steps and require a strong understanding of core calculus concepts, like integration and substitution. This exercise demonstrates the thought process behind solving integrals by using substitutions. It's a classic example of how calculus problems require a combination of algebraic manipulation and strategic thinking.
Here’s a brief breakdown:

• Identify the trigonometric identities and their potential to simplify the integral.
• Use substitution to transform the integral into an easier form to compute.
• Perform the integration and back-substitute to the original variable.

This clear, methodical approach not only helps in solving the problem but also reinforces critical calculus skills. By repeatedly practicing exercises like this one, you enhance your ability to tackle a wide range of calculus problems. These exercises build intuition for when and how to apply different integration techniques, making them an essential part of any calculus curriculum.