Problem 317 - Calculus Volume 1

Question

For the following exercises, find the definite or indefinite integral. $$\int_{0}^{\pi / 4} \tan x d x$$

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{1}{2} \ln 2 \).

Step 1: Recognize Integration Method

Given the integral \( \int_{0}^{\pi / 4} \tan x \, dx \), we need to apply a suitable integration technique. The function \( \tan x \) does not have a straightforward integral, but we recognize that \( \tan x = \frac{\sin x}{\cos x} \), suggesting a substitution method is appropriate.

Step 2: Apply Substitution

To integrate \( \tan x \), substitute \( u = \cos x \), then \( du = -\sin x \, dx \). Hence, \( \sin x \, dx = -du \) and the integral now becomes:\[\int \frac{\sin x}{\cos x} \, dx = \int -\frac{1}{u} \, du\].

Step 3: Integrate Simplified Expression

The integral \( \int -\frac{1}{u} \, du \) is straightforward and results in:\[-\ln |u| + C\]Substituting back for \( u = \cos x \), we have:\[-\ln | \cos x | + C\].

Step 4: Evaluate Definite Integral from \(0\) to \(\frac{\pi}{4}\)

Substitute the bounds into the antiderivative. Compute:\[-\left[ \ln | \cos x | \right]_{0}^{\pi/4} = -\ln | \cos (\pi/4) | + \ln | \cos 0 | \].

Step 5: Calculate Trigonometric Values and Simplify

Calculate \( \cos \left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \) and \( \cos(0) = 1 \). Substitute to get:\[-(\ln | \frac{\sqrt{2}}{2} | - \ln | 1 | ) = -\ln \left( \frac{\sqrt{2}}{2} \right) = \ln 2 - \frac{1}{2} \ln 2 = \frac{1}{2} \ln 2 \].

Key Concepts

Substitution Method

The substitution method is a common technique for simplifying integrals, especially when dealing with trigonometric functions. It is useful for transforming messy functions into ones that are easier to integrate.
In this exercise, the original function, \( \tan x = \frac{\sin x}{\cos x} \), doesn't lend itself to immediate integration using standard rules. Here, we cleverly choose \( u = \cos x \) as our substitution variable. The rationale behind this is to convert the function into a simpler form that we can easily integrate.
This substitution requires rewriting \( \sin x \, dx \) in terms of \( du \). Since \( du = -\sin x \, dx \), we can express \( \sin x \, dx = -du \). Thus, the integral \( \int \frac{\sin x}{\cos x} \, dx \) becomes \( \int -\frac{1}{u} \, du \). This transformation reduces the complexity of the integral dramatically, allowing us to proceed with standard integration methods.

Trigonometric Functions

Trigonometric functions are fundamental in calculus. They often appear in integrals and can make direct integration challenging.
Here, we encounter the trigonometric function \( \tan x \), which is expressed as \( \frac{\sin x}{\cos x} \). This identity helps us use the substitution method effectively by recognizing that dividing sine by cosine leads directly into an integration pathway that involves substitution.
When working with trigonometric functions in integrals, it’s critical to understand these basic identities:

• \( \sin^2 x + \cos^2 x = 1 \)
• \( \tan x = \frac{\sin x}{\cos x} \)
• \( \cot x = \frac{\cos x}{\sin x} \)

In this way, shifting between different trigonometric functions can simplify the integration process.

Antiderivative

Finding the antiderivative, or the indefinite integral, is a key step in solving an integral. It tells us which function, when differentiated, will give back our original function.
In this example, once we substitute and simplify the integral, we find ourselves with \( \int -\frac{1}{u} \, du \), whose antiderivative is \( -\ln |u| + C \). Switching back from \( u \) to \( \cos x \), we obtain an expression in terms of \( x \).
Computing definite integrals then involves evaluating this antiderivative between given limits. For the range from 0 to \( \frac{\pi}{4} \), we calculate the expression \( -\ln | \cos x | \) at these boundaries. The subtraction gives us the definite integral's value.\( \cos \left(\frac{\pi}{4}\right) \) and \( \cos(0) \) provide specific values used in the evaluation:

• \( \cos \left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \)
• \( \cos(0) = 1 \)

Using these calculations, the definite integral of \( \tan x \) from 0 to \( \frac{\pi}{4} \) is ultimately simplified to \( \frac{1}{2} \ln 2 \). This process illustrates the role of antiderivatives in determining the area under the curve specified by our limits.