Problem 113 - Calculus for Biology and Medicine

Question

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

Step-by-Step Solution

Answer

\(\frac{1}{2} \ln(2)\)

Step 1: Find the Indefinite Integral

To evaluate the definite integral, we first find the indefinite integral of \( \tan(2x) \). The integral of \( \tan(u) \) is \( -\ln|\cos(u)| \). Here, \( u = 2x \) so we need to apply the substitution method. Let \( u = 2x \), hence \( du = 2dx \) or \( dx = \frac{1}{2}du \). The indefinite integral becomes: \[ \int \tan(2x) \, dx = \frac{1}{2} \int \tan(u) \, du = -\frac{1}{2} \ln|\cos(u)| + C \]Substitute back \( u = 2x \): \[ -\frac{1}{2} \ln|\cos(2x)| + C \]

Step 2: Apply the Fundamental Theorem of Calculus

To find the definite integral from 0 to \( \frac{\pi}{6} \), we apply the Fundamental Theorem of Calculus. This involves evaluating the antiderivative at the upper limit and subtracting the value of the antiderivative at the lower limit:\[ \int_{0}^{\pi/6} \tan(2x) \, dx = \left[ -\frac{1}{2} \ln|\cos(2x)| \right]_{0}^{\pi/6} \]Calculate each part of the evaluation by substituting the bounds.

Step 3: Evaluate at Upper and Lower Bounds

Substitute the upper bound \( x = \frac{\pi}{6} \):\[ -\frac{1}{2} \ln|\cos(\frac{2\pi}{6})| = -\frac{1}{2} \ln|\cos(\frac{\pi}{3})| = -\frac{1}{2} \ln\left( \frac{1}{2} \right) \]Substitute the lower bound \( x = 0 \):\[ -\frac{1}{2} \ln|\cos(0)| = -\frac{1}{2} \ln(1) = -\frac{1}{2} \times 0 = 0 \]

Step 4: Calculate the Definite Integral

Now, calculate the difference between the evaluated values: \[ \int_{0}^{\pi/6} \tan(2x) \, dx = \left( -\frac{1}{2} \ln\left( \frac{1}{2} \right) \right) - (0) \]This simplifies to: \[ \int_{0}^{\pi/6} \tan(2x) \, dx = \frac{1}{2} \ln(2) \].

Key Concepts

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is a key principle linking the concept of derivatives and integrals. It essentially provides a powerful connection between these two areas of calculus. Here's how it works: when you have the indefinite integral of a function, the theorem tells us that we can use it to find the definite integral, which represents the total area under the curve of a function over a specified interval.
To apply this theorem, you:

• Take the antiderivative, or the indefinite integral.
• Evaluate it at the upper limit of the interval.
• Subtract the value when evaluated at the lower limit.

This process converts differentiation back to the function from which the derivative was derived, and subsequently allows us to compute definite integrals. In the example with the integral of \( \int_{0}^{\pi / 6} \tan (2 x) \, dx \), this allowed us to find the exact area using the bound evaluation of the antiderivative.

Substitution Method

The Substitution Method is a technique used to make integrals easier to solve. It involves changing the variable of integration to simplify the expression, making it potentially easier to evaluate the integral.
Here's how you use substitution:

• Identify a substitution \( u = g(x) \) that simplifies the integral.
• Compute the differential \( du = g'(x)dx \).
• Rewrite the integral in terms of \( u \) and \( du \).
• Integrate with respect to \( u \).
• Finally, return to the original variable by back-substituting \( u \) with \( g(x) \).

In the original problem, setting \( u = 2x \) and therefore \( du = 2dx \) simplifies \( \tan(2x)dx \) to \( \frac{1}{2}\tan(u)du \), underscoring why \( u \)-substitution is an effective strategy.

Indefinite Integral

An indefinite integral, often written as an integral without bounds, represents a family of functions and includes a constant of integration \( C \). It is effectively the reverse process of differentiation, and is sometimes called an antiderivative.
When you're working with indefinite integrals:

• You are essentially finding what function has a given derivative.
• The result includes \( C \), since differentiation of a constant results in zero, making multiple solutions possible.

In our specific example, finding the indefinite integral \( \int \tan(2x) \, dx \) led us to the antiderivative \( -\frac{1}{2} \ln|\cos(2x)| + C \). This forms the basis for calculating definite integrals later, providing a generalized solution over an interval.