Question

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

Short answer

Question: Write down the expressions for the integrals of \(e^{ax}\sin BX\) and \(e^{ax}\cos bx\). Answer: The expression for the integral of \(e^{ax}\sin bx\) is: \(\int e^{ax}\sin bx\,dx=\frac{e^{ax}(a\sin bx-b\cos bx)}{a^2+b^2}+C\) The expression for the integral of \(e^{ax}\cos bx\) is: \(\int e^{ax}\cos bx\,dx=\frac{e^{ax}(a\cos bx+b\sin bx)}{a^2+b^2}+C\)

Step-by-step solution

Identify u and dv for the first integral

For the first integral, \(\int e^{ax}\sin bx\,dx\), we choose: \(u = \sin bx\) and \(dv = e^{ax}\,dx\) Now, differentiate \(u\) to get \(du\) and integrate \(dv\) to get \(v\): \(du = b\cos bx\,dx\) \(v = \frac{1}{a}e^{ax}\)

Apply integration by parts for the first integral

Using the integration by parts formula on the first integral, we get: $$\int e^{ax}\sin bx\,dx = \frac{1}{a}e^{ax}\sin bx - \int \frac{1}{a}e^{ax} b\cos bx\,dx$$ For this new integral, choose: \(u = \cos bx\) and \(dv = \frac{1}{a}e^{ax}\,dx\) Differentiate \(u\) to get \(du\) and integrate \(dv\) to get \(v\): \(du = -b\sin bx\,dx\) \(v = \frac{1}{a^2}e^{ax}\)

Apply integration by parts again for the first integral

Using the integration by parts formula on the new integral, we get: $$\int e^{ax}\sin bx\,dx = \frac{1}{a}e^{ax}\sin bx - \left(\frac{1}{a^2}e^{ax}\cos bx - \int \frac{1}{a^2}e^{ax} b\sin bx\,dx\right)$$ Rearranging the equation, we have: $$\int e^{ax}\sin bx\,dx = \frac{1}{a}e^{ax}\sin bx - \frac{1}{a^2}e^{ax}\cos bx + \int \frac{1}{a^2}e^{ax} b\sin bx\,dx$$

Solve for the remaining integral in the first equation

Now we are left with an integral that is only a constant multiple away from our original integral. To solve, multiply both sides by the constant: $$\left(1-\frac{b^2}{a^2}\right)\int e^{ax}\sin bx\,dx = \frac{e^{ax}}{a}(a\sin bx - b\cos bx)$$ Now, divide by the constant term: $$\int e^{ax}\sin bx\,dx=\frac{e^{ax}(a\sin bx-b\cos bx)}{a^2+b^2}+C$$ Now we will work on the second integral.

Identify u and dv for the second integral

For the second integral, \(\int e^{ax}\cos bx\,dx\), we choose: \(u = \cos bx\) and \(dv = e^{ax}\,dx\) Now, differentiate \(u\) to get \(du\) and integrate \(dv\) to get \(v\): \(du = -b\sin bx\,dx\) \(v = \frac{1}{a}e^{ax}\)

Apply integration by parts for the second integral

Using the integration by parts formula for the second integral, we get: $$\int e^{ax}\cos bx\,dx = \frac{1}{a}e^{ax}\cos bx - \int \frac{1}{a}e^{ax} (-b)\sin bx\,dx$$ For this new integral, choose: \(u = \sin bx\) and \(dv = \frac{1}{a}e^{ax}\,dx\) Differentiate \(u\) to get \(du\) and integrate \(dv\) to get \(v\): \(du = b\cos bx\,dx\) \(v = \frac{1}{a^2}e^{ax}\)

Apply integration by parts again for the second integral

Using the integration by parts formula on the new integral, we get: $$\int e^{ax}\cos bx\,dx = \frac{1}{a}e^{ax}\cos bx + \left(\frac{1}{a^2}e^{ax}\sin bx - \int \frac{1}{a^2}e^{ax} b\cos bx\,dx\right)$$ Rearranging the equation, we have: $$\int e^{ax}\cos bx\,dx = \frac{1}{a}e^{ax}\cos bx + \frac{1}{a^2}e^{ax}\sin bx - \int \frac{1}{a^2}e^{ax} b\cos bx\,dx$$

Solve for the remaining integral in the second equation

Now we are left with an integral that is only a constant multiple away from our original integral. To solve, multiply both sides by the constant: $$\left(1-\frac{b^2}{a^2}\right)\int e^{ax}\cos bx\,dx = \frac{e^{ax}}{a}(a\cos bx + b\sin bx)$$ Now, divide by the constant term: $$\int e^{ax}\cos bx\,dx=\frac{e^{ax}(a\cos bx+b\sin bx)}{a^2+b^2}+C$$ We have successfully derived both integrals using integration by parts.

Key concepts

Exponential Functions

Exponential functions play a crucial role in calculus and integration by parts, especially when they multiply with other functions. These functions have the general form of \( e^{ax} \), where \( e \) is the base of natural logarithms, and \( a \) is a constant. Exponential functions are essential because they appear naturally in various mathematical scenarios, such as growth and decay processes. To understand their significance in integration by parts, it's vital to know that exponential functions often simplify because their derivatives and integrals result in another exponential function. This property is useful when combined with other functions, like trigonometric functions, as it allows the original problem to be broken down into more manageable parts.

• Exponential functions grow or decay at a rate proportional to their current value.
• They have the special property that the slope of the tangent line to the graph of the function at any point is equal to the value of the function at that point.
• In integration, their constant derivatives make them preferable for certain strategies, like integration by parts.

Trigonometric Functions

Trigonometric functions such as \( \sin bx \) and \( \cos bx \) are periodic functions that are often involved in integration by parts problems. These functions are useful in modeling wave-like phenomena and oscillatory behaviors. They have specific relationships and identities that are often used to simplify integration steps. The derivative and integral of sine and cosine functions swap between each other, with signs changing, allowing them to be paired conveniently with exponential functions in integration scenarios. When exploiting integration by parts, trigonometric functions tend to "cycle" back, making them interdependent and thus helping to construct formulas through the process of repeated integration by parts.

• Sine and cosine have derivatives that result in the other function: \( \frac{d}{dx} \sin bx = b\cos bx \) and \( \frac{d}{dx} \cos bx = -b\sin bx \).
• In integration by parts, when trigonometric functions are paired with exponential functions, they help in achieving a balance or symmetry that simplifies integration.
• They often involve identities and transformations that aid in integrating complex expressions.

Definite Integrals

Definite integrals describe the net area under the curve of a function over an interval \([a, b]\). They provide insight into the accumulation of quantities and are often used to compute physical quantities like work or probability. In the context of integration by parts, definite integrals help establish fixed boundaries for evaluating the integral of a function. With definite integrals, the boundaries do not change the structure of the integration by parts technique, but they do result in evaluating the resulting expression at these points when computing the integral results. This evaluation adds a layer of complexity as one must keep track of these boundary values throughout the integration process.

• Definitive calculations require evaluating the integrated function at the upper and lower limits and subtracting these values.
• They are represented by the formula \( \int_{a}^{b} f(x) \, dx \), providing a finite result contrasted with indefinite integrals that include the constant \( C \).
• When using integration by parts on definite integrals, apply the limits after finding the integrated form.

Mathematical Derivation

Mathematical derivation involves systematically developing a formula or statement based on logical reasoning and mathematical principles. This process is essential in integration by parts, as it transforms a complex integral into simpler terms or an analytical form. Mathematical derivation goes beyond just finding the integral; it's about understanding the underlying connections and patterns within functions. Integration by parts is a powerful tool to perform these derivations, particularly for integrals that combine exponential and trigonometric functions. The derivation process in this method often involves:

• Selecting appropriate functions for \( u \) and \( dv \) such that the integral simplifies when using \( \int u \, dv = uv - \int v \, du \).
• Executing multiple iterations of integration by parts when needed, until reaching a solvable point or a repetitive cycle.
• Rearranging results to isolate the original integral, effectively deriving a new formula or confirming the derived equation.

The derivation showcases ingenuity in mathematical problem-solving, highlighting patterns that can be generalized or applied in broader contexts.