Problem 12 - Biocalculus Calculus for the Life Sciences

Question

Evaluate the integral. \(\int e^{-\theta} \cos 2 \theta d \theta\)

Step-by-Step Solution

Answer

\( \int e^{-\theta} \cos 2\theta \, d\theta = -\frac{1}{3} e^{-\theta} ( \cos 2\theta - \sin 2\theta ) + C \).

Step 1: Identify the Integration Method

The integral \( \int e^{-\theta} \cos 2\theta \, d\theta \) requires the use of integration by parts, as both \( e^{-\theta} \) and \( \cos 2\theta \) are standard functions for which integration by parts can be effectively applied.

Step 2: Select \( u \) and \( dv \) for Integration by Parts

Choose \( u = \cos 2\theta \) which will give \( du = -2\sin 2\theta \, d\theta \). Then, choose \( dv = e^{-\theta} \, d\theta \), which means that \( v = -e^{-\theta} \) after integration.

Step 3: Apply the Integration by Parts Formula

Using the integration by parts formula \( \int u \, dv = uv - \int v \, du \), substitute the values: \[ \int e^{-\theta} \cos 2\theta \, d\theta = (-e^{-\theta} \cos 2\theta) - \int (-e^{-\theta})(-2\sin 2\theta) \, d\theta \].

Step 4: Simplify and Repeat Integration by Parts

The integral becomes \( -e^{-\theta} \cos 2\theta + 2\int e^{-\theta} \sin 2\theta \, d\theta \). We need to use integration by parts on the second integral. Set \( u = \sin 2\theta \) and \( dv = e^{-\theta} d\theta \), giving \( du = 2\cos 2\theta \, d\theta \), \( v = -e^{-\theta} \).

Step 5: Integrate the Second Integral

Apply integration by parts: \[ 2 \left( -e^{-\theta} \sin 2\theta + \int e^{-\theta} \cdot 2\cos 2\theta \, d\theta \right) \]. Substitute this back into our original equation.

Step 6: Substitute and Solve for Final Result

Upon substitution, it results again in the original integral. Therefore, consider it as \( I = \int e^{-\theta} \cos 2\theta \, d\theta \) leading to \( I = -e^{-\theta} ( \cos 2\theta - \sin 2\theta ) - 2I \). Solve for \( I \): \[ 3I = -e^{-\theta} ( \cos 2\theta - \sin 2\theta ) \].

Step 7: Simplify Solution

Divide by 3 to isolate \( I \): \[ I = -\frac{1}{3} e^{-\theta} ( \cos 2\theta - \sin 2\theta ) + C \] where \( C \) is the constant of integration.

Key Concepts

Exponential Functions

Exponential functions, like the one in our integral, exhibit continuous and rapid growth or decay. In the math world, these are often denoted by functions like \( e^{x} \), where \( e \) is Euler's number, approximately 2.71828. These functions are unique because their rate of growth is proportional to their current value, which is why they are immensely useful in modeling growth and decay processes, such as populations or radioactive decay.
When working with integrals involving exponential functions, their distinctive property is that the derivative or integral of an exponential function with a linear argument is proportional to the function itself. For example, the derivative of \( e^{x} \) is still \( e^{x} \). This makes them incredibly predictable and easy to work with, especially when mixed with other functions like trigonometric and polynomial functions.
In integration, exponential functions often pair with other functions in problems requiring integration by parts. This method leverages their predictable behavior to deal with more complex integrations.

Trigonometric Functions

Trigonometric functions, such as \( \cos x \) and \( \sin x \), are periodic functions that arise from the study of angles and the lengths of sides of triangles. They have key properties derived from the unit circle which gives them unique behavior such as periodicity, meaning they repeat their values in regular intervals.
These functions are immensely useful in a variety of fields such as physics, engineering, and even in financial mathematics, where they can model waves, oscillations and other periodic phenomena. For integration, trigonometric functions often require special methods, as their integrations can lead to other trigonometric functions or even result in logarithmic functions when paired with tangents.
In the integral \( \int e^{-\theta} \cos 2\theta \, d\theta \), \( \cos 2\theta \) operates as a trigonometric function that, when paired with an exponential function, usually requires careful selection of integration techniques, like integration by parts, for a simplified solution.

Definite and Indefinite Integrals

In calculus, integrals come in two main forms: definite and indefinite. An indefinite integral, such as \( \int e^{-\theta} \cos 2\theta \, d\theta \), represents a family of functions and includes the constant of integration \( C \), which accounts for the constant nature of the function when derivating. It answers the question of what function, if differentiated, would provide the original function under the integral sign.
On the other hand, definite integrals calculate a number and represent the area under a curve between two points. They solve the question of how much total accumulated quantity exists across that interval. For indefinite integrals, the focus is on generalizing a function's antiderivative, conveying all potential solutions before specific bounds are applied.
When solving integrals, being aware of whether you are dealing with a definite or indefinite integral influences your approach and helps determine the steps that follow. In our exercise, because it's presented as an indefinite integral, it leads us to introduce a constant \( C \) for the solution.