Problem 284 - Calculus Volume 1

Question

In the following exercises, use a suitable change of variables to determine the indefinite integral. $$ \int \cos \theta(1-\cos \theta)^{99} \sin \theta d \theta $$

Step-by-Step Solution

Answer

The indefinite integral is \( \frac{(1 - \cos \theta)^{100}}{100} - \frac{(1 - \cos \theta)^{101}}{101} + C \).

Step 1: Identify a Suitable Change of Variables

The integral involves both \( \cos \theta \) and \((1 - \cos \theta)^{99}\). To simplify the expression, we use a substitution \( u = 1 - \cos \theta \). This substitution will help simplify the power term in the integrand.

Step 2: Differentiate to Find du

By differentiating the substitution \( u = 1 - \cos \theta \), we get \[ du = \sin \theta \, d\theta \]This indicates \( \sin \theta \, d\theta \) is indeed present, making our substitution approach suitable.

Step 3: Rewrite the Integral in Terms of u

Substitute \( u = 1 - \cos \theta \) and \( du = \sin \theta \, d\theta \) into the integral:\[ \int \cos \theta (1 - \cos \theta)^{99} \sin \theta \, d\theta = \int \cos \theta \cdot u^{99} \cdot \sin \theta \, d\theta \]Since \( \cos \theta = 1 - u \), the integral becomes:\[ \int (1-u) u^{99} \, du \]

Step 4: Simplify and Integrate

Expand the integral:\[ \int ((1 \cdot u^{99}) - (u \cdot u^{99})) \, du = \int (u^{99} - u^{100}) \, du \]Integrate term by term:\[ \int u^{99} \, du = \frac{u^{100}}{100} \]\[ \int u^{100} \, du = \frac{u^{101}}{101} \]Combine the results:\[ \int u^{99} - u^{100} \, du = \frac{u^{100}}{100} - \frac{u^{101}}{101} + C \]

Step 5: Substitute Back to Original Variables

Substitute \( u = 1 - \cos \theta \) back into the antiderivative:\[ \int \cos \theta (1 - \cos \theta)^{99} \sin \theta \, d\theta = \frac{(1 - \cos \theta)^{100}}{100} - \frac{(1 - \cos \theta)^{101}}{101} + C \]

Key Concepts

Change of Variables

The technique of 'change of variables' in integration, often referred to as substitution, is a powerful method to simplify integrals. Essentially, you're transforming the variable of integration into another variable that makes the integral easier to solve. In the context of this exercise, we were dealing with \[ \int \cos \theta (1 - \cos \theta)^{99} \sin \theta \, d \theta \]. The term \((1 - \cos \theta)^{99}\) suggested using a substitution that would simplify managing such a high power. By setting \[ u = 1 - \cos \theta \],we change the variable from \( \theta \) to \( u \). This turns a potentially complicated problem into a more straightforward polynomial integration, which is much easier to handle.

Integration by Substitution

Integration by substitution is essentially the reverse of the chain rule from differentiation. It allows us to simplify complex integrals by substituting part of the integral with a single variable. Here's how it worked in our example:

• First, we chose \( u = 1 - \cos \theta \) as our substitution to simplify the power involved.
• Next, we found \( du \) by differentiating \( u \), which gives \( du = \sin \theta \, d\theta \).
• This expression for \( du \) matched the \( \sin \theta \, d\theta \) part in the integral, confirming our substitution.

With this substitution, we rewrote the integral in terms of \( u \), which made the integration much simpler. It’s a clever trick to turn something complex into a manageable form.

Trigonometric Identities

Trigonometric identities are invaluable tools in calculus for simplifying expressions that involve trigonometric functions. These identities reveal interconnections between trigonometric functions. For instance:- In the exercise, the identity used was implicit through the substitution \( u = 1 - \cos \theta \), which separates terms involving sine and cosine efficiently.Knowing these identities can save time and effort when facing integrals involving trigonometric functions. They provide alternate ways to view and approach problems, often simplifying them and making them solvable with basic integration techniques. Understanding and using these identities effectively can transform a complex-looking integral into a straightforward problem.