Problem 23 - Calculus Early Transcendentals: Pearson New International Edition

Question

Perform the indicated integrations. $$ \int \tan ^{5}\left(\frac{\theta}{2}\right) d \theta $$

Step-by-Step Solution

Answer

The integral of \( \tan^5\left(\frac{\theta}{2}\right) d\theta \) is \( \frac{\tan^4\left(\frac{\theta}{2}\right)}{2} - \tan^2\left(\frac{\theta}{2}\right) + C \).

Step 1: Rewrite the Integral using Trigonometric Identities

To solve the integral \( \int \tan^5\left(\frac{\theta}{2}\right) d\theta \), we can start by rewriting \( \tan\left(\frac{\theta}{2}\right) \) in terms of \( \sin \) and \( \cos \) using the identity: \( \tan\left(\frac{\theta}{2}\right) = \frac{1 - \cos(\theta)}{\sin(\theta)} \). This will help us to express the tangent in terms of sine and cosine for easier integration.

Step 2: Use the Substitution

Let \( u = \tan\left(\frac{\theta}{2}\right) \). Then \( \sin(\theta) = \frac{2u}{1+u^2} \) and \( \cos(\theta) = \frac{1-u^2}{1+u^2} \). Also, \( d\theta = \frac{2 du}{1+u^2} \). Substitute these into the integral to transform it.

Step 3: Convert the Original Integral

Substituting, we have \( \int u^5 \cdot \frac{2}{1+u^2} du \). Now the integral in terms of \( u \) becomes \( 2 \int \frac{u^5}{1+u^2} du \).

Step 4: Use Integration by Parts if Necessary

Consider using integration by parts for \( 2 \int \frac{u^5}{1+u^2} du \). Alternatively, simplify using polynomial long division if necessary, before attempting other integration techniques.

Step 5: Simplify and Integrate

Rewrite \( \frac{u^5}{1+u^2} = u^3 - u \) by polynomial division. This makes the integration process simpler: \( 2 \int (u^3 - u) du \).

Step 6: Compute the Integral

Now integrate term-by-term: \[ 2 \times \left( \frac{u^4}{4} - \frac{u^2}{2} \right) = \frac{u^4}{2} - u^2. \]

Step 7: Substitute Back for \( \theta\)

Replace \( u \) with \( \tan\left(\frac{\theta}{2}\right) \) to transform back into the original variable: \( \frac{\tan^4\left(\frac{\theta}{2}\right)}{2} - \tan^2\left(\frac{\theta}{2}\right) + C \), where \( C \) is the constant of integration.

Key Concepts

Trigonometric Identities

Trigonometric identities are essential tools in calculus, especially for tackling integrals involving trigonometric functions. In the given exercise, the identity \( \tan\left(\frac{\theta}{2}\right) = \frac{1 - \cos(\theta)}{\sin(\theta)} \) is crucial for simplifying the integral. Understanding how to express complex trigonometric expressions in simpler forms is key. This identity allows us to rewrite the tangent half-angle in terms of sine and cosine, which can be more accessible for integration. Simplifying expressions through such identities can make solving integrals much more straightforward. When solving an integrals strategy, like substituting trigonometric identities, can simplify the equation and make integration feasible. Keep these identities handy as they often provide the stepping stones to solving more complex integrals.

Substitution Method

The substitution method in integration is somewhat akin to the reverse of the chain rule in differentiation. It simplifies complex integrals by introducing a new variable. In the exercise, we use the substitution \( u = \tan\left(\frac{\theta}{2}\right) \). With this substitution, we change the variable of integration and modify the integral to make it more manageable. The differential \( d\theta \) is also converted using \( d\theta = \frac{2 du}{1+u^2} \). Through substitution, the complexity of the trigonometric integral is reduced considerably, transforming it into an algebraic one.The main goal of substitution is to transform the integral into a form that is easier to evaluate. It’s one of the most powerful techniques in calculus, providing a bridge from challenging integrals to simpler algebraic formats.

Integration by Parts

Integration by parts is a powerful technique central to solving integrals, particularly when the integrals involve the product of functions. It is based on the product rule for derivatives and is expressed as: \[ \int u \, dv = uv - \int v \, du \]In the context of the exercise, while it's suggested to consider integration by parts, the polynomial long division actually simplifies it further. However, understanding integration by parts as a concept prepares you for tackling integrals of products of functions.When faced with an integral of the form \(2 \int \frac{u^5}{1+u^2} du \), integration by parts might initially seem useful. Often, breaking down the integral into parts and applying the rule can help in evaluating it if simpler methods, like polynomial division, don't suffice.

Polynomial Division

Polynomial division simplifies complex fractions in integrals. In this exercise, we use polynomial division to simplify \( \frac{u^5}{1+u^2} \) into \( u^3 - u \).This method divides the polynomial in the numerator by the polynomial in the denominator, much like long division in arithmetic. Once simplified, we can integrate each term separately, which makes the integration much more straightforward.Applying polynomial division before integration reduces complexity and allows us to handle simpler polynomial expressions. This strategy, in many cases, avoids more complicated integration techniques and directly gives us a manageable expression to integrate.