Problem 85 - Calculus Single Variable

Question

Suppose that \(a\) and \(b\) are constants, not both \(0 .\) Derive the formula $$ \int e^{a x} \cos (b x) d x=\frac{e^{a x}(a \cos (b x)+b \sin (b x))}{a^{2}+b^{2}}+C . $$

Step-by-Step Solution

Answer

Integrate by parts twice and solve algebraically to derive the formula.

Step 1: Use Integration by Parts Formula

Recall the integration by parts formula: \(\int u \, dv = uv - \int v \, du\). We need to choose \(u\) and \(dv\) for our integral \(\int e^{ax} \cos(bx) \, dx\). Let \(u = \cos(bx)\) and \(dv = e^{ax} \, dx\). Then differentiate \(u\) to get \(du = -b \sin(bx) \, dx\), and integrate \(dv\) to get \(v = \frac{1}{a} e^{ax}\).

Step 2: Apply Integration by Parts

Apply the integration by parts formula: \(\int e^{ax} \cos(bx) \, dx = \frac{1}{a} e^{ax} \cos(bx) - \int \left(-b \sin(bx)\right) \frac{1}{a} e^{ax} \, dx\). Simplify this to \(\int e^{ax} \cos(bx) \, dx = \frac{1}{a} e^{ax} \cos(bx) + \frac{b}{a} \int e^{ax} \sin(bx) \, dx\).

Step 3: Integrate Resultant Integral by Parts Again

Now, apply integration by parts to \(\int e^{ax} \sin(bx) \, dx\). Choose \(u = \sin(bx)\) and \(dv = e^{ax} \, dx\). Thus, \(du = b \cos(bx) \, dx\) and \(v = \frac{1}{a} e^{ax}\). So, we have \(\int e^{ax} \sin(bx) \, dx = \frac{1}{a} e^{ax} \sin(bx) - \frac{b}{a} \int e^{ax} \cos(bx) \, dx\).

Step 4: Combine Integrals and Solve for Original Integral

Substitute the expression for \(\int e^{ax} \sin(bx) \, dx\) back into the equation from Step 2: \(\int e^{ax} \cos(bx) \, dx = \frac{1}{a} e^{ax} \cos(bx) + \frac{b}{a} \left( \frac{1}{a} e^{ax} \sin(bx) - \frac{b}{a} \int e^{ax} \cos(bx) \, dx \right)\).

Step 5: Simplify and Solve Algebraically

Distribute the \(\frac{b}{a}\) and simplify: \(\int e^{ax} \cos(bx) \, dx = \frac{1}{a} e^{ax} \cos(bx) + \frac{b}{a^2} e^{ax} \sin(bx) - \frac{b^2}{a^2} \int e^{ax} \cos(bx) \, dx\). Let \(I = \int e^{ax} \cos(bx) \, dx\). Solve for \(I\) to obtain: \(I + \frac{b^2}{a^2} I = \frac{1}{a} e^{ax} \cos(bx) + \frac{b}{a^2} e^{ax} \sin(bx)\), which simplifies to \(I (1 + \frac{b^2}{a^2}) = \frac{e^{ax}(a \cos(bx) + b \sin(bx))}{a^2}\).

Step 6: Solve for I and Finalize the Solution

Multiply both sides by \(\frac{a^2}{a^2 + b^2}\) to solve for \(I\): \(I = \frac{e^{ax} (a \cos(bx) + b \sin(bx))}{a^2 + b^2}\). Thus, the integral \(\int e^{ax} \cos(bx) \, dx = \frac{e^{ax} (a \cos(bx) + b \sin(bx))}{a^2 + b^2} + C\).

Key Concepts

Indefinite Integrals

Indefinite integrals, often referred to as antiderivatives, are functions that are derived by reversing the process of differentiation. Unlike definite integrals, indefinite integrals do not have upper and lower limits. Instead, they include a constant of integration, denoted as "C". This constant is essential because functions that differ by a constant have the same derivative, making it impossible to identify the original function exactly.

When dealing with indefinite integrals, the aim is to find a function whose derivative matches the given integrand. The process involves relying on known rules and techniques, including substitution or integration by parts, as seen in the exercise. Integration by parts is particularly vital when the integrand is a product of two functions.

• Integration by parts formula: \(\int u \, dv = uv - \int v \, du\).
• The choice of \(u\) and \(dv\) is strategic, often making differentiation of \(u\) (resulting in \(du\)) simpler while integrating \(dv\).

This formula helps in systematically breaking down complex integrals, as displayed in the exercise solution.

Exponential Functions

Exponential functions are a class of functions in the form \(e^{ax}\), where \(e\) is the base of natural logarithms and \(a\) is a constant. Their unique property is that the function’s rate of growth is proportional to its current value, meaning that differentiating or integrating \(e^{ax}\) results in very predictable forms.

• When differentiated with respect to \(x\), \(e^{ax}\) remains an exponential function: \(\frac{d}{dx} e^{ax} = ae^{ax}\).
• Similarly, when integrated, an exponential function continues to appear as \(\int e^{ax} \, dx = \frac{1}{a} e^{ax} + C\).

This predictable behavior is especially beneficial. It allows us to focus on applying other integration techniques, such as integration by parts, without being concerned about the fundamental structure changing.
In the exercise above, exponential functions are used alongside trigonometric functions to illustrate how such problems can be approached and simplified through systematic integration techniques.

Trigonometric Functions

Trigonometric functions like \(\sin(bx)\) and \(\cos(bx)\) frequently appear in mathematical problems involving waves, oscillations, and rotations. They have periodic properties that can be neatly manipulated using calculus.

In integration, these functions behave in ways parallel to their derivatives. For instance, the integral of \(\cos(bx)\) with respect to \(x\) is \(\frac{1}{b} \sin(bx)\), and similarly, the integral of \(\sin(bx)\) is \(-\frac{1}{b} \cos(bx)\) plus a constant.

• Trigonometric integrals often require methods like integration by parts or trigonometric identities to simplify them.
• They can appear in tandem with other types of functions, like exponential functions, requiring a blended approach to integration.

In this exercise, the interplay between trigonometric and exponential functions showcases the necessity to comprehend both tendencies to navigate through the process of integration successfully. By dissecting the integral piece by piece, and utilizing methods like integration by parts, intricate expressions become more manageable and solvable.