Question

Two useful exponential integrals Use integration by parts to derive the following formulas for real numbers $a$ and $b$. $$\begin{array}{l}\int e^{ax}\sin bxdx=\frac{e^{ax}(a\sin bx-b\cos bx)}{a^2+b^2}+C\\ \int e^{ax}\cos bxdx=\frac{e^{ax}(a\cos bx+b\sin bx)}{a^2+b^2}+C\end{array}$$

Short answer

In this exercise, we derived the formulas for the exponential integrals involving exponential functions and trigonometric functions using integration by parts. Following a step-by-step approach, we assigned functions to be integrated and differentiated, calculated the intermediate steps, and finally derived the formulas as requested. The resulting formulas for the integrals are: $$\int e^{ax}\sin bxdx=\frac{e^{ax}(a\sin bx-b\cos bx)}{a^2+b^2}+C_1$$ and $$\int e^{ax}\cos bxdx=\frac{e^{ax}(a\cos bx+b\sin bx)}{a^2+b^2}+C_2$$

Step-by-step solution

Integration by parts

In this step, we will assign the functions to be integrated and differentiated. For the first integral, we assign $u=e^{ax}$ and $dv=\sin bxdx$. Similarly, for the second integral, we assign $u=e^{ax}$ and $dv=\cos bxdx$.

Finding du and v for the first integral

Now, we need to differentiate u and integrate dv for the first integral. Differentiating u, we get $du=a{e}^{ax}dx$. Integrating dv, we have $v=\int \sin bxdx=-\frac{1}{b}\cos bx$.

Applying integration by parts to the first integral

Now, let's apply the integration by parts formula to the first integral: $\int udv=uv-\int vdu$. Here, $u=e^{ax}$, $dv=\sin bxdx$, $du=a{e}^{ax}dx$, and $v=-\frac{1}{b}\cos bx$. So, we have: $$\int e^{ax}\sin bxdx=-\frac{1}{b}{e}^{ax}\cos bx-\int (-\frac{a}{b}{e}^{ax}\cos bx)dx$$

Applying integration by parts again for the first integral

Let's apply integration by parts again for the integral on the right-hand side. This time, we have $u=e^{ax}$ and $dv=\cos bxdx$. Differentiating u, we get $du=a{e}^{ax}dx$. Integrating dv, we have $v=\int \cos bxdx=\frac{1}{b}\sin bx$. Applying the formula, we get: $$\int e^{ax}\cos bxdx=\frac{1}{b}{e}^{ax}\sin bx-\int (\frac{a}{b}{e}^{ax}\sin bx)dx$$ Substituting this result into the equation from step 3, we have: $$\int e^{ax}\sin bxdx=-\frac{1}{b}{e}^{ax}\cos bx+\frac{a}{b^2}{e}^{ax}\sin bx-\frac{a^2}{b^2}\int e^{ax}\sin bxdx$$

Solving for the first integral

Now let's get the $\int e^{ax}\sin bxdx$ term on one side of the equation and factor it out: $$\int e^{ax}\sin bxdx(1+\frac{a^2}{b^2})=\frac{e^{ax}}{b^2}(-b\cos bx+a\sin bx)$$ Dividing both sides by $(1+\frac{a^2}{b^2})$, we get the final result for the first integral: $$\int e^{ax}\sin bxdx=\frac{e^{ax}(a\sin bx-b\cos bx)}{a^2+b^2}+C_1$$

Finding du and v for the second integral, and applying integration by parts

For the second integral, we will follow similar steps as above. We have differentiated u before and got $du=a{e}^{ax}dx$. Integrating dv, we get $v=\int \cos bxdx=\frac{1}{b}\sin bx$. Applying integration by parts formula, we have: $$\int e^{ax}\cos bxdx=\frac{1}{b}{e}^{ax}\sin bx-\int (\frac{a}{b}{e}^{ax}\sin bx)dx$$

Applying integration by parts again for the second integral

Let's apply integration by parts again for the integral on the right-hand side of the equation from step 6. We know $du=a{e}^{ax}dx$ and we get $v=\int \sin bxdx=-\frac{1}{b}\cos bx$. Applying the formula, we get: $$\int e^{ax}\sin bxdx=-\frac{1}{b}{e}^{ax}\cos bx-\int (-\frac{a}{b}{e}^{ax}\cos bx)dx$$ Substituting this result into the equation from step 6, we have: $$\int e^{ax}\cos bxdx=\frac{1}{b}{e}^{ax}\sin bx+\frac{a}{b^2}{e}^{ax}\cos bx-\frac{a^2}{b^2}\int e^{ax}\cos bxdx$$

Solving for the second integral

Now let's get the $\int e^{ax}\cos bxdx$ term on one side of the equation and factor it out: $$\int e^{ax}\cos bxdx(1+\frac{a^2}{b^2})=\frac{e^{ax}}{b^2}(b\sin bx+a\cos bx)$$ Dividing both sides by $(1+\frac{a^2}{b^2})$, we get the final result for the second integral: $$\int e^{ax}\cos bxdx=\frac{e^{ax}(a\cos bx+b\sin bx)}{a^2+b^2}+C_2$$ Now we have derived the given formulas for the exponential integrals using integration by parts.

Key concepts

Integration by Parts

Understanding integration by parts is essential for solving integrals that involve products of functions. It's a technique that is based on the product rule for differentiation and provides a way to integrate the product of two functions. The formula for integration by parts is expressed as $$\int udv=uv-\int vdu.$$This formula requires us to choose which part of our integrand to differentiate $u$ and which part to integrate $dv$. The correct choice can simplify the problem significantly.
When applying integration by parts to exponential integrals, such as $e^{ax}\times \text{sin}(bx)$ or $e^{ax}\times \text{cos}(bx)$, the exponential function is often a good choice for $u$, since its derivative remains an exponential function, thus keeping the form suitable for subsequent steps. The trigonometric function is then $dv$, which cycles back after differentiation and integration (sine changes to cosine or vice versa), making it easier to establish a relationship between the initial integral and the one derived from the integration by parts.

Integrating Exponential Functions

Working with integrating exponential functions often appears in calculus problems due to their prevalence in natural phenomena like growth and decay. The general form of an exponential function is $e^{ax}$, where $a$ is a constant, and $e$ is the base of the natural logarithm. The integral of an exponential function is straightforward: $$\int e^{ax}dx=\frac{e^{ax}}{a}+C,$$ where $C$ is the constant of integration.
However, when combined with other functions, such as trigonometric functions, the integration becomes more challenging. The exercise given involves such a combination, and it shows that integrating exponential functions with other types of functions may sometimes involve multiple iterations of techniques like integration by parts.

Solving Integral Equations

An integral equation is an equation where an unknown function appears under an integral sign. Solving integral equations is a prominent part of applied mathematics, appearing in various science and engineering problems. The exercise showcases a method for solving integral equations involving exponential and trigonometric functions.

The key is to express the problem in a solvable form, often requiring manipulation and algebraic operations. After applying integration by parts, one might obtain an integral equation that includes the original integral. This calls for algebraic techniques to isolate the original integral and solve for it, as seen in the steps that involve grouping terms and factoring the integral out. The art of solving integral equations lies in recognizing patterns and strategically applying calculus and algebraic methods.

Calculus Problem-Solving

Effective calculus problem-solving requires a deep understanding of various mathematical concepts and the ability to apply appropriate techniques to specific problems. In the context of the provided exercise, problem-solving involves recognizing that integration by parts is the correct approach, selecting the appropriate $u$ and $dv$, and handling the algebraic manipulation after applying the technique.
To successfully solve calculus problems, it's also crucial to:

• Understand the underlying principles of the problem.
• Identify the type of functions involved.
• Determine the best strategy based on the problem's characteristics.
• Perform algebraic manipulations accurately.
• Keep track of constants of integration.

By mastering these skills through practice and exposure to different types of problems, one develops the problem-solving agility necessary to tackle a wide range of calculus challenges.