Problem 24 - Calculus Early Transcendentals: Pearson New International Edition

Question

$$ \text { use integration by parts to evaluate each integral. } $$ $$ \int_{\pi / 6}^{\pi / 4} x \sec ^{2} x d x $$

Step-by-Step Solution

Answer

\( \frac{\pi}{4} - \frac{\pi}{6\sqrt{3}} + \frac{1}{2} \ln \frac{2}{3} \)

Step 1: Identify Components for Integration by Parts

Integration by parts is based on the formula \( \int u \, dv = uv - \int v \, du \). In this exercise, we will choose \( u = x \) and \( dv = \sec^2 x \, dx \). This means we need to compute \( du \) and \( v \).

Step 2: Compute the Derivatives and Integrals

Given \( u = x \), then \( du = dx \). For \( dv = \sec^2 x \, dx \), integrating gives \( v = \tan x \).

Step 3: Apply the Integration by Parts Formula

Using the formula: \( \int u \, dv = uv - \int v \, du \), substitute the known values to get: \[ \int x \sec^2 x \, dx = x \tan x - \int \tan x \, dx \].

Step 4: Evaluate the Remaining Integral

The next task is to evaluate \( \int \tan x \, dx \). We know \( \int \tan x \, dx = -\ln |\cos x| + C \). Thus, substituting back, we get: \[ x \tan x + \ln |\cos x| \].

Step 5: Evaluate the Definite Integral

Now evaluate the expression \( x \tan x + \ln |\cos x| \) from \( \pi/6 \) to \( \pi/4 \). Calculate: \[ \left( \frac{\pi}{4} \tan \frac{\pi}{4} + \ln |\cos \frac{\pi}{4}| \right) - \left( \frac{\pi}{6} \tan \frac{\pi}{6} + \ln |\cos \frac{\pi}{6}| \right) \].

Step 6: Simplify the Results

Compute each term: \( \tan \frac{\pi}{4} = 1 \) and \( \cos \frac{\pi}{4} = \frac{\sqrt{2}}{2} \); \( \tan \frac{\pi}{6} = \frac{1}{\sqrt{3}} \) and \( \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2} \). Substitute these into the expression: \[ \frac{\pi}{4}(1) + \ln \frac{\sqrt{2}}{2} - \left( \frac{\pi}{6} \frac{1}{\sqrt{3}} + \ln \frac{\sqrt{3}}{2} \right) \].

Step 7: Final Simplified Result

Simplify further to arrive at the final result: \( \frac{\pi}{4} - \frac{\pi}{6\sqrt{3}} + \frac{1}{2} \ln 2 - \frac{1}{2} \ln 3 \).

Key Concepts

Definite Integrals

Definite integrals are a key part of calculus, especially when you want to find the net area under a curve over a specific interval. They go beyond just measuring area; they also capture the accumulation of quantities and the net change over an interval.

In the context of integration by parts, definite integrals involve applying limits to the evaluated result. The process usually starts with finding the indefinite integral (the antiderivative) and then using the Fundamental Theorem of Calculus. For the given exercise, once the indefinite integral is acquired,

• apply the upper and lower limits directly to the function we've integrated.
• Subtract the result of the lower limit evaluation from the result of the upper limit evaluation.

This process helps transform an abstract calculus operation into a concrete numerical answer.

Trigonometric Functions

Trigonometric functions are central to calculus, especially when dealing with integrals involving angles, such as the one in our exercise. Functions like secant \((\sec x)\) and tangent \((\tan x)\) appear regularly, and their integrals often require careful handling.

The integral of trigonometric functions can sometimes be tricky compared to polynomials or exponential functions. For example, integrating \( \sec^2 x \) directly yields \( \tan x \), a useful derivative fact often used in calculus. It's important to remember these fundamental relationships:

• The derivative of \( \tan x \) is \( \sec^2 x \).
• The antiderivative of \( \sec^2 x \) is \( \tan x \).
• When integrating trigonometric functions, be attentive to identities and relationships that can simplify the process.

These relationships make the solving of integrals a bit smoother and more intuitive. Understanding these shortcuts can significantly enhance your problem-solving efficiency.

Calculus Methods

Integration by parts is a powerful calculus method for solving integrals that are products of functions. It can be especially useful when straightforward integration isn't feasible. The method is derived from the product rule for differentiation and is written as:
\[ \int u \, dv = uv - \int v \, du \]

Choosing how to assign \( u \) and \( dv \) is crucial and can affect the simplicity of the process. In typical practice:

• Choose \( u \) to be a function that simplifies when differentiated.
• Choose \( dv \) such that its integral, \( v \), is straightforward.

In our exercise, selecting \( u = x \) and \( dv = \sec^2 x \, dx \) leverages these choices effectively. Calculating \( du = dx \) and \( v = \tan x \) brings us directly to the solution without overly complicating the integral.

After applying the integration by parts formula, you'll sometimes encounter another integral to solve, as we did with \( \int \tan x \, dx \). Using known integrals or integration techniques like substitution or further integration by parts will simplify this step, bringing you closer to the solution.