Problem 16 - University Calculus: Early Transcendentals

Question

Evaluate the integrals. $$\int_{0}^{\pi / 6}(\sec x+\tan x)^{2} d x$$

Step-by-Step Solution

Answer

The integral evaluates to \(\sqrt{3} - 1\).

Step 1: Understand the Integral

We need to evaluate the definite integral \( \int_{0}^{\pi / 6} (\sec x + \tan x)^2 \, dx \). It's useful to start by expanding \((\sec x + \tan x)^2\) to simplify the integral.

Step 2: Expand the Expression

The expression \((\sec x + \tan x)^2\) can be expanded as follows: \[ (\sec x + \tan x)^2 = \sec^2 x + 2 \sec x \tan x + \tan^2 x \].

Step 3: Simplify Using Identities

Recall the trigonometric identities: \( \sec^2 x = 1 + \tan^2 x \) and the derivative of \( \sec x \) is \( \sec x \tan x \). So, \( \sec^2 x + 2 \sec x \tan x + \tan^2 x = (1 + \tan^2 x) + 2 \sec x \tan x \).

Step 4: Recognize Derivative Form

Notice that the derivative of \( \sec x + \tan x \) is \( \sec x \tan x + \sec^2 x \). Hence, the original function's derivative we expanded matches with our expanded integral form partially.

Step 5: Compute the Integral

Since the integrand \( \sec^2 x + 2 \sec x \tan x + \tan^2 x \) corresponds to the derivative of \( \sec x + \tan x \), the integral simplifies to \( [\sec x + \tan x]_{0}^{\pi / 6} \).

Step 6: Evaluate at the Limits

Now evaluate: \( [\sec x + \tan x]_{0}^{\pi / 6} = (\sec(\pi/6) + \tan(\pi/6)) - (\sec(0) + \tan(0)) \).

Step 7: Substitute Trigonometric Values

Substitute the trigonometric values: \( \sec(\pi/6) = 2/\sqrt{3} \) and \( \tan(\pi/6) = 1/\sqrt{3} \), giving \( [2/\sqrt{3} + 1/\sqrt{3}] - [1 + 0] \).

Step 8: Simplify the Expression

Simplifying the expression: \( 3/\sqrt{3} - 1 = \sqrt{3} - 1 \).

Step 9: Final Result

The final result of the integral is \( \sqrt{3} - 1 \).

Key Concepts

Trigonometric Functions

When dealing with definite integrals, understanding trigonometric functions plays a crucial role. Trigonometric functions such as sine \(\sin\), cosine \(\cos\), tangent \(\tan\), and secant \(\sec\) are foundational in calculus. In our exercise, we focus on \(\sec x\) and \(\tan x\).

• Secant, \(\sec x\), is the reciprocal of cosine \(\cos x\). It expresses as \(\sec x = \frac{1}{\cos x}\).
• Tangent, \(\tan x\), is the ratio of sine to cosine: \(\tan x = \frac{\sin x}{\cos x}\).

You'll notice these functions were utilized to form the expression \(\sec x + \tan x\), which was then squared and expanded. Recognizing their reciprocal and ratio forms can help demystify the manipulation you're performing. Remember, knowing how these functions relate to each other through identities and reciprocal laws is essential in calculus.

Integration Techniques

To solve definite integrals, understanding integration techniques is essential. In this exercise, expanding the expression \(\sec x + \tan x\) to \(\sec^2 x + 2 \sec x \tan x + \tan^2 x\) is key. Here's why:

• Expanding helps break down complex expressions into a form easier to integrate.
• After expanding, observe the known derivatives—this is your shortcut to evaluate the integral without detailed computation.

After expanding, we leveraged the knowledge of derivatives. This allows us to align the integrand with the derivative of a function (in this case, \(\sec x + \tan x\)). Such recognition often simplifies the integration process significantly by reducing it to evaluating a function at the given limits.

Calculus Identities

Calculus identities form the backbone of simplifying integrals and derivatives. In our task, two key trigonometric identities were crucial:

• \(\sec^2 x = 1 + \tan^2 x\) helps relate \(\sec^2 x\) and \(\tan^2 x\) directly through a simple identity. This identity was instrumental in simplifying and relating expressions.
• The derivative connection: noticing that the derivative of \(\sec x + \tan x\) is precisely part of the expanded integrand, allowed for simplification directly to evaluating a known antiderivative.

By understanding such identities and their role in transforming integrals, your ability to solve integrals will improve. These identities are especially valuable in reducing expressions and facilitating substitutions, thereby streamlining both differentiation and integration processes.