Problem 85 - Thomas Calculus

Question

a. Express \(\int \tan ^{3} \theta d \theta\) in terms of \(\int \tan \theta d \theta\) . Then evaluate \(\int \tan ^{3} \theta d \theta \cdot\left(\text {Hint} : \tan ^{2} \theta=\sec ^{2} \theta-1 .\right)\) b. Express \(\int \tan ^{5} \theta d \theta\) in terms of \(\int \tan ^{3} \theta d \theta\) c. Express \(\int \tan ^{7} \theta d \theta\) in terms of \(\int \tan ^{5} \theta d \theta\) d. Express \(\int \tan ^{2 k+1} \theta d \theta,\) where \(k\) is a positive integer, in terms of \(\int \tan ^{2 k-1} \theta d \theta\)

Step-by-Step Solution

Answer

a) \( \frac{1}{2} \tan^2 \theta + \ln |\cos \theta| + C \); b), c), d) use the recursive pattern: \( \int \tan^{2k+1} \theta \, d\theta = \frac{1}{2k} \tan^{2k} \theta - \int \tan^{2k-1} \theta \, d\theta. \)

Step 1: Use Trigonometric Identity

In part a, we are asked to express \( \int \tan^3 \theta \, d\theta \) in terms of \( \int \tan \theta \, d\theta \). We start by using the hint given: \( \tan^2 \theta = \sec^2 \theta - 1 \). Substitute this identity into the integral: \[ \int \tan^3 \theta \, d\theta = \int \tan \theta (\tan^2 \theta) \, d\theta = \int \tan \theta (\sec^2 \theta - 1) \, d\theta. \]

Step 2: Split the Integral

The integral can now be split into two separate integrals: \[ \int \tan \theta \sec^2 \theta \, d\theta - \int \tan \theta \, d\theta. \] We recognize that \( \int \tan \theta \sec^2 \theta \, d\theta \) can be simplified by using a substitution method.

Step 3: Simplify Using Substitution

Let \( u = \tan \theta \), then \( du = \sec^2 \theta \, d\theta \). Substitute this into the first integral: \[ \int u \, du = \frac{1}{2} u^2 + C_1 = \frac{1}{2} \tan^2 \theta + C_1. \] Thus, \( \int \tan \theta \sec^2 \theta \, d\theta = \frac{1}{2} \tan^2 \theta + C_1. \)

Step 4: Combine and Express the Integral

Now, substitute back the result from Step 3 into our original split integrals: \[ \int \tan^3 \theta \, d\theta = \frac{1}{2} \tan^2 \theta - \int \tan \theta \, d\theta. \] This expresses \( \int \tan^3 \theta \, d\theta \) in terms of \( \int \tan \theta \, d\theta \).

Step 5: Evaluate the Integral

Recall that \( \int \tan \theta \, d\theta = -\ln |\cos \theta| + C_2 \). Thus, the integral becomes: \[ \int \tan^3 \theta \, d\theta = \frac{1}{2} \tan^2 \theta + \ln |\cos \theta| + C. \]

Step 6: Generalize the Identity for Odd Powers

In parts b, c, and d, we seek patterns. For \( \int \tan^{2k+1} \theta \, d\theta \), observe that: \[ \int \tan^{2k+1} \theta \, d\theta = \int \tan^{2k-1} \theta \tan^2 \theta \, d\theta = \int \tan^{2k-1} \theta (\sec^2 \theta - 1) \, d\theta. \] Split this into: \[ \int \tan^{2k-1} \theta \sec^2 \theta \, d\theta - \int \tan^{2k-1} \theta \, d\theta. \]

Step 7: Establish Pattern for \(\tan^{2k+1} \theta\)

Applying substitution on the first part where \( u = \tan \theta \), we get \( du = \sec^2 \theta \, d\theta \). Thus, \( \int \tan^{2k-1} \theta \sec^2 \theta \, d\theta \to \int u^{2k-1} \, du = \frac{u^{2k}}{2k} + C_3 \). Hence \[ \int \tan^{2k+1} \theta \, d\theta = \frac{1}{2k} \tan^{2k} \theta - \int \tan^{2k-1} \theta \, d\theta. \]

Step 8: Conclude with General Equation

The pattern identified will help in writing any integral of \( \tan^{2k+1} \theta \) in terms of the integral of \( \tan^{2k-1} \theta \): \[ \int \tan^{2k+1} \theta \, d\theta = \frac{1}{2k} \tan^{2k} \theta - \int \tan^{2k-1} \theta \, d\theta. \] This recursive relationship answers parts b, c, and d.

Key Concepts

Trigonometric Integrals

Trigonometric integrals often involve integrating functions that include trigonometric expressions like sine, cosine, or tangent. These can be challenging without the proper techniques.
In the given exercise, we are dealing with an integral involving the tangent function to an odd power. To solve it efficiently, we need to apply trigonometric identities. One such useful identity is - \( \tan^2 \theta = \sec^2 \theta - 1 \), which helps us to simplify the given trigonometric integral. Using this identity, we can rewrite higher powers of tangent in terms of the secant function. This transformation is crucial because it allows us to break a complex integral into simpler parts. Trigonometric integrals often require creativity and a good understanding of these identities to solve effectively.

Substitution Method

The substitution method is a technique used to simplify complex integrals by changing variables. It is especially useful when dealing with integrals of trigonometric functions.
In the exercise, the substitution method was applied to the integral \( \int \tan \theta \sec^2 \theta \, d\theta \). By letting \( u = \tan \theta \), the differential becomes \( du = \sec^2 \theta \, d\theta \). This changes the integral into a simpler form: - \( \int u \, du \). This can then be easily integrated to yield \( \frac{1}{2} u^2 + C \). Substitution not only simplifies the integration process but also provides a clearer path to the solution.

Integration by Parts

Integration by parts is another powerful technique that is particularly useful in solving integrals involving products of functions. It relies on the formula: - \( \int u \, dv = uv - \int v \, du \).In the context of the original exercise, by manipulating the tangential powers using identities and substitution, we essentially prepare the integral to potentially benefit from integration by parts in further explorations.Though not directly employed in the steps provided, this method is pivotal in integral calculus and often complements the substitution method. When faced with trigonometric integrals, understanding when and how to apply integration by parts can unlock solutions to seemingly difficult problems.

Recursive Patterns in Integration

Recursive patterns in integration arise when we can express an integral in terms of another integral of a similar form. This is a powerful method for tackling integrals of higher powers, as seen in parts b, c, and d of the exercise.By expressing \( \int \tan^{2k+1} \theta \, d\theta \) in terms of \( \int \tan^{2k-1} \theta \, d\theta \): - \( \int \tan^{2k+1} \theta \, d\theta = \frac{1}{2k} \tan^{2k} \theta - \int \tan^{2k-1} \theta \, d\theta \), we create a recursive relationship.
This pattern allows us to solve integrals with higher odd powers of tangent in a step-by-step manner, using solutions to simpler problems. Recursive methods are invaluable in integral calculus, providing structured routes to solve complex problems by reducing them to simpler base cases.