Problem 32 - Calculus of a Single Variable

Question

Finding an Indefinite Integral of a Trigonometric Function In Exercises \(31-40\) , find the indefinite integral. $$ \int \tan 5 \theta d \theta $$

Step-by-Step Solution

Answer

The indefinite integral of the trigonometric function \( \int \tan5 \theta d \theta \) is \( -\frac{1}{5} \ln |\cos 5 \theta | + C \)

Step 1: Rewrite tangent

The first step is to rewrite the \(\tan 5 \theta\) as \(\frac{\sin 5 \theta}{\cos 5 \theta} \). This is done because the integral of tangent is more easily evaluated when it's written as a ratio of sine to cosine. Our integral now reads: \[\int \frac{\sin 5 \theta}{\cos 5 \theta} d \theta\]

Step 2: Substitution

The second step is to use a substitution. Here, let \( u = \cos 5 \theta \). Find the differential of \( u \), which is \( du = -5 \sin 5 \theta d \theta \). Our new expression becomes: \[\int \frac{-1}{5u} du \]

Step 3: Integrate

Now, integrate the expression \(\int \frac{-1}{5u} du\), which is \(-\frac{1}{5} \ln |u| + C \). Now, replace \( u \) with \(\cos 5 \theta \) to put it back in terms of \( \theta \) . So the final answer would be:\[-\frac{1}{5} \ln |\cos 5\theta| + C\]

Key Concepts

Trigonometric Functions

Trigonometric functions like sine, cosine, and tangent are essential in many branches of mathematics and engineering. They relate the angles of a right triangle to the lengths of its sides and are periodic functions, meaning they repeat their values in regular intervals. This property makes them especially useful when solving problems involving periodic phenomena, such as waves and oscillations.

In the given exercise, we focus on the tangent function, denoted as \( \tan 5\theta \). The tangent function can be tricky to integrate in its standard form. A helpful strategy is to express it in terms of sine and cosine. Specifically, \( \tan \theta = \frac{\sin \theta}{\cos \theta} \). This transformation often simplifies the integration process.

When dealing with trigonometric functions, remember these useful identities:

• \( \sin^2 \theta + \cos^2 \theta = 1 \)
• \( \sec^2 \theta = 1 + \tan^2 \theta \)
• \( \tan 2\theta = \frac{2 \tan \theta}{1 - \tan^2 \theta} \)

Using these identities can make manipulating and simplifying trigonometric expressions much easier.

Substitution Method

The substitution method, also known as "u-substitution," is a classic technique in calculus used to simplify integration. The idea is straightforward: you replace a part of the integrand with a variable, typically \( u \), that simplifies the expression.

In our exercise, we substituted \( u = \cos 5\theta \). This choice was strategic because the derivative, \( du = -5 \sin 5\theta \, d\theta \), directly appeared in the integral's sine component. This substitution transformed the original integral into a simpler form, \( \int \frac{-1}{5u} \, du \).

The basic steps for substitution are:

• Identify a portion of the integrand to replace with \( u \).
• Differentiate \( u \) to find \( du \).
• Rewrite the integral using \( u \) and \( du \).
• Perform the integration in terms of \( u \).
• Finally, substitute back the original variable to express the result in terms of the original variable.

Substitution is powerful for functions that aren't straightforward to integrate directly, especially when dealing with angles or exponents.

Integration Techniques

Integration techniques encompass a variety of methods for finding the integral of functions, which is essentially the reverse process of differentiation. For the integral of \( \tan 5\theta \), the primary technique used was substitution, as previously explained. However, understanding various integration techniques can prepare you for tackling a wide array of problems.

Other common techniques include:

• **Integration by Parts:** Useful for products of functions, based on the product rule for differentiation.
• **Partial Fractions:** Decomposes rational functions into simpler fractions that are easier to integrate.
• **Trigonometric Substitution:** Useful for integrals involving square roots of quadratic expressions.

In our example, after performing the substitution, integrating the transformed expression \( \int \frac{-1}{5u} \, du \) results in using the formula \( \int \frac{1}{x} \, dx = \ln |x| + C \). This commonly used result appears frequently in logarithmic integration scenarios.

Practicing different techniques expands your mathematical toolkit, allowing you to tackle more complex integrals with confidence.