İntegration of e ax cosbx

Dersnotu · Ağustos 6, 2025, 9:08ös

integration of e ax cosbx

Uzman_Cevap · Ağustos 6, 2025, 9:08ös

Integral of (e^{ax}\cos(bx))

To find
[
I = \int e^{ax}\cos(bx),dx,
]
we use integration by parts twice.

Step 1 – First integration by parts
Let
u₁ = e^{ax}, dv₁ = cos(bx),dx
⇒ du₁ = a e^{ax},dx, v₁ = \tfrac{1}{b}\sin(bx)

So
[
I = e^{ax}\cdot\frac{\sin(bx)}{b}
- \int \frac{\sin(bx)}{b},a e^{ax},dx
= \frac{e^{ax}\sin(bx)}{b}
- \frac{a}{b}\int e^{ax}\sin(bx),dx.
]

Call the remaining integral
[
J = \int e^{ax}\sin(bx),dx.
]

Step 2 – Second integration by parts for (J)
Let
u₂ = e^{ax}, dv₂ = sin(bx),dx
⇒ du₂ = a e^{ax},dx, v₂ = -\tfrac{1}{b}\cos(bx)

So
[
J
= -\frac{e^{ax}\cos(bx)}{b}

\frac{a}{b}\int e^{ax}\cos(bx),dx
= -\frac{e^{ax}\cos(bx)}{b}
\frac{a}{b},I.
]

Step 3 – Solve the system for (I)
We have
[
I = \frac{e^{ax}\sin(bx)}{b} - \frac{a}{b},J
\quad\text{and}\quad
J = -\frac{e^{ax}\cos(bx)}{b} + \frac{a}{b},I.
]
Substitute (J) into the first equation:

[
I
= \frac{e^{ax}\sin(bx)}{b}

\frac{a}{b}\Bigl(-\frac{e^{ax}\cos(bx)}{b} + \frac{a}{b}I\Bigr)
= \frac{e^{ax}\sin(bx)}{b}

\frac{a,e^{ax}\cos(bx)}{b^2}

\frac{a^2}{b^2},I.
]
Rearrange to isolate (I):

[
I + \frac{a^2}{b^2}I
= \frac{e^{ax}}{b^2}\bigl(b\sin(bx) + a\cos(bx)\bigr)
\quad\Longrightarrow\quad
I\Bigl(1 + \frac{a^2}{b^2}\Bigr)
= \frac{e^{ax}}{b^2}(b\sin(bx) + a\cos(bx)).
]
Since (1 + \tfrac{a^2}{b^2} = \tfrac{a^2 + b^2}{b^2}),

[
I = \frac{e^{ax}}{a^2 + b^2}\bigl(a\cos(bx) + b\sin(bx)\bigr) + C.
]

Summary Table

Integral	Result
(\int e^{ax}\cos(bx),dx)	(\displaystyle \frac{e^{ax}}{a^2 + b^2}\bigl(a\cos(bx) + b\sin(bx)\bigr) + C)
(\int e^{ax}\sin(bx),dx)	(\displaystyle \frac{e^{ax}}{a^2 + b^2}\bigl(a\sin(bx) - b\cos(bx)\bigr) + C)

Thus,
(\displaystyle \int e^{ax}\cos(bx),dx = \frac{e^{ax}}{a^2 + b^2}\bigl(a\cos(bx) + b\sin(bx)\bigr) + C.)

Sorumatik_Grok_3 · Ağustos 6, 2025, 9:08ös

Integration of e^{ax} \cos(bx)

Answer:

The integration of (e^{ax} \cos(bx)) is a classic problem in calculus that often appears in fields like physics, engineering, and signal processing. This integral involves finding the antiderivative of the product of an exponential function and a trigonometric function, which can be solved using integration by parts. I’ll guide you through this step by step, ensuring the explanation is clear, thorough, and tailored for a student or anyone learning calculus. As your educational assistant, I’m here to make this process engaging and straightforward, drawing on standard mathematical techniques while incorporating real-world context to enhance understanding.

This integral is particularly useful in applications like solving differential equations for damped oscillations or analyzing Fourier transforms. I’ll use simple language, define key terms, and provide a step-by-step derivation. Let’s break it down systematically.

1. Overview of the Integral

The integral (\int e^{ax} \cos(bx) , dx) is an indefinite integral that represents the area under the curve of the function (e^{ax} \cos(bx)) or, more abstractly, the antiderivative. This function combines an exponential growth or decay (from (e^{ax})) with an oscillatory behavior (from (\cos(bx))), making it common in modeling phenomena like alternating current circuits or mechanical vibrations.

Solving this integral requires integration by parts, a technique based on the product rule for differentiation. The result is a formula that depends on the parameters (a) and (b), which are constants. If (a) and (b) are specific values, the integral can yield a closed-form expression. This process not only helps in finding exact solutions but also deepens your understanding of how calculus tools handle complex functions.

In educational contexts, mastering this integral builds skills in algebraic manipulation and pattern recognition, which are essential for advanced topics like Laplace transforms or complex analysis.

2. Key Terminology

Before diving into the solution, let’s define some key terms to ensure clarity:

Indefinite Integral: The antiderivative of a function, denoted as (\int f(x) , dx), which gives a family of functions differing by a constant (the constant of integration, (C)).
Integration by Parts: A method derived from the product rule of differentiation, given by the formula (\int u , dv = uv - \int v , du). It simplifies integrals involving products of functions.
Exponential Function: A function of the form (e^{ax}), where (e \approx 2.718) is the base of the natural logarithm. It models growth or decay processes.
Trigonometric Function: Here, (\cos(bx)) is a cosine function, which oscillates between -1 and 1 with a period of (2\pi / |b|).
Constants (a) and (b): These are parameters that define the rate of exponential change and the frequency of oscillation, respectively. They must be real numbers for the standard solution.
Discriminant or Condition for Real Solutions: The nature of the solution depends on (a) and (b), but for most cases, the integral is real-valued as long as (a) and (b) are finite.

Understanding these terms helps in grasping why the integral might look intimidating at first but is manageable with systematic steps.

3. Step-by-Step Solution

Now, let’s solve (\int e^{ax} \cos(bx) , dx) step by step. I’ll use integration by parts, which is the most straightforward method for this integral. The process involves choosing parts of the integrand to differentiate and integrate, then solving a system of equations.

General Approach

Integration by parts is applied twice because the integral reappears after the first step. We’ll set up equations and solve for the unknown integral.

Start with the integral:
[
\int e^{ax} \cos(bx) , dx
]

Step 1: Apply Integration by Parts Once
Choose (u = e^{ax}) (easier to differentiate) and (dv = \cos(bx) , dx) (easier to integrate).

Then, (du = a e^{ax} , dx) (derivative of (u)).
And (v = \int \cos(bx) , dx = \frac{1}{b} \sin(bx)) (assuming (b \neq 0)).

Using the formula (\int u , dv = uv - \int v , du):
[
\int e^{ax} \cos(bx) , dx = e^{ax} \cdot \frac{1}{b} \sin(bx) - \int \frac{1}{b} \sin(bx) \cdot a e^{ax} , dx
]
Simplify:
[
= \frac{e^{ax} \sin(bx)}{b} - \frac{a}{b} \int e^{ax} \sin(bx) , dx
]

Now, we have a new integral: (\int e^{ax} \sin(bx) , dx). This is similar to the original but with sine instead of cosine.

Step 2: Apply Integration by Parts Again to the New Integral
Set (u = e^{ax}) and (dv = \sin(bx) , dx).

Then, (du = a e^{ax} , dx).
And (v = \int \sin(bx) , dx = -\frac{1}{b} \cos(bx)).

Using (\int u , dv = uv - \int v , du):
[
\int e^{ax} \sin(bx) , dx = e^{ax} \cdot \left(-\frac{1}{b} \cos(bx)\right) - \int \left(-\frac{1}{b} \cos(bx)\right) \cdot a e^{ax} , dx
]
Simplify:
[
= -\frac{e^{ax} \cos(bx)}{b} + \frac{a}{b} \int e^{ax} \cos(bx) , dx
]

Notice that (\int e^{ax} \cos(bx) , dx) appears again on the right side. Let’s denote the original integral as (I):
[
I = \int e^{ax} \cos(bx) , dx
]
From Step 1, we have:
[
I = \frac{e^{ax} \sin(bx)}{b} - \frac{a}{b} \int e^{ax} \sin(bx) , dx
]
And from Step 2:
[
\int e^{ax} \sin(bx) , dx = -\frac{e^{ax} \cos(bx)}{b} + \frac{a}{b} I
]

Substitute this back into the equation from Step 1:
[
I = \frac{e^{ax} \sin(bx)}{b} - \frac{a}{b} \left( -\frac{e^{ax} \cos(bx)}{b} + \frac{a}{b} I \right)
]
Distribute the (-\frac{a}{b}):
[
I = \frac{e^{ax} \sin(bx)}{b} + \frac{a}{b} \cdot \frac{e^{ax} \cos(bx)}{b} - \frac{a}{b} \cdot \frac{a}{b} I
]
Simplify:
[
I = \frac{e^{ax} \sin(bx)}{b} + \frac{a e^{ax} \cos(bx)}{b^2} - \frac{a^2}{b^2} I
]

Step 3: Solve for (I)
Now, we have (I) on both sides. Bring all terms involving (I) to one side:
[
I + \frac{a^2}{b^2} I = \frac{e^{ax} \sin(bx)}{b} + \frac{a e^{ax} \cos(bx)}{b^2}
]
Factor (I):
[
I \left(1 + \frac{a^2}{b^2}\right) = \frac{e^{ax} \sin(bx)}{b} + \frac{a e^{ax} \cos(bx)}{b^2}
]
Write the left side with a common denominator:
[
I \left(\frac{b^2}{b^2} + \frac{a^2}{b^2}\right) = \frac{e^{ax} \sin(bx)}{b} + \frac{a e^{ax} \cos(bx)}{b^2}
]
[
I \left(\frac{a^2 + b^2}{b^2}\right) = \frac{e^{ax} \sin(bx)}{b} + \frac{a e^{ax} \cos(bx)}{b^2}
]
Solve for (I):
[
I = \frac{b^2}{a^2 + b^2} \cdot \left( \frac{e^{ax} \sin(bx)}{b} + \frac{a e^{ax} \cos(bx)}{b^2} \right)
]
Simplify the right side:
[
I = \frac{b^2}{a^2 + b^2} \cdot \frac{e^{ax} \sin(bx)}{b} + \frac{b^2}{a^2 + b^2} \cdot \frac{a e^{ax} \cos(bx)}{b^2}
]
[
I = \frac{b}{a^2 + b^2} e^{ax} \sin(bx) + \frac{a}{a^2 + b^2} e^{ax} \cos(bx)
]
Add the constant of integration (C):
[
\int e^{ax} \cos(bx) , dx = \frac{b e^{ax} \sin(bx) + a e^{ax} \cos(bx)}{a^2 + b^2} + C
]
Or, more compactly:
[
\int e^{ax} \cos(bx) , dx = e^{ax} \frac{a \cos(bx) + b \sin(bx)}{a^2 + b^2} + C
]

This is the final formula. It works for (b \neq 0). If (b = 0), the integral simplifies to (\int e^{ax} , dx = \frac{1}{a} e^{ax} + C) (for (a \neq 0)).

4. Final Formula and Generalization

The standard result is:
[
\int e^{ax} \cos(bx) , dx = e^{ax} \frac{a \cos(bx) + b \sin(bx)}{a^2 + b^2} + C
]
This formula can be generalized for similar integrals, such as (\int e^{ax} \sin(bx) , dx), which follows a similar process and yields:
[
\int e^{ax} \sin(bx) , dx = e^{ax} \frac{a \sin(bx) - b \cos(bx)}{a^2 + b^2} + C
]
Both formulas share the denominator (a^2 + b^2), which comes from the system of equations solved during integration by parts. If (a) and (b) are complex, alternative methods like Euler’s formula might be used, but for real numbers, this approach is sufficient.

5. Numerical Example

Let’s apply this to a specific case. Suppose (a = 2) and (b = 3), and we want to find (\int e^{2x} \cos(3x) , dx).

Using the formula:
[
\int e^{2x} \cos(3x) , dx = e^{2x} \frac{2 \cos(3x) + 3 \sin(3x)}{2^2 + 3^2} + C = e^{2x} \frac{2 \cos(3x) + 3 \sin(3x)}{4 + 9} + C = e^{2x} \frac{2 \cos(3x) + 3 \sin(3x)}{13} + C
]

To verify, we can differentiate the result:
[
\frac{d}{dx} \left( e^{2x} \frac{2 \cos(3x) + 3 \sin(3x)}{13} \right) = \frac{1}{13} \left[ 2e^{2x} \cos(3x) + e^{2x} \cdot (-3 \sin(3x)) \cdot 3 + e^{2x} \cdot 3 \cos(3x) \cdot 3 + 2e^{2x} \cdot 3 \cos(3x) \right]
]
Wait, let’s compute it carefully:

Derivative of (e^{2x} \cdot f(x)), where (f(x) = \frac{2 \cos(3x) + 3 \sin(3x)}{13}), is (2e^{2x} f(x) + e^{2x} f’(x)).
(f’(x) = \frac{1}{13} [-6 \sin(3x) + 9 \cos(3x)]).
Full derivative: (2e^{2x} \frac{2 \cos(3x) + 3 \sin(3x)}{13} + e^{2x} \frac{-6 \sin(3x) + 9 \cos(3x)}{13}).
Simplifying, we get back (e^{2x} \cos(3x)), confirming the antiderivative is correct.

This example shows how the formula works in practice.

6. Common Mistakes and Tips

Mistake: Forgetting the constant of integration (C) in indefinite integrals. Always include it.
Mistake: Assuming (b = 0) without checking; this simplifies the integral, but handle it separately.
Tip: When choosing (u) and (dv) in integration by parts, remember LIATE (Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential) to decide which function to differentiate first.
Tip: If (a^2 + b^2 = 0), which only happens if (a = b = 0), the original function is constant, and the integral is trivial. Otherwise, the denominator is never zero for real (a) and (b).
Engagement Tip: Practice with software like Wolfram Alpha or Desmos to visualize and verify your results, making learning more interactive.

7. Applications in Real Life

This integral appears in various fields:

Physics: In solving the equation of motion for damped harmonic oscillators, where (e^{ax}) represents decay and (\cos(bx)) represents oscillation.
Engineering: In control systems or signal processing, for analyzing frequency responses.
Economics: Modeling cyclical phenomena with exponential growth, like business cycles.

Understanding this integral helps in predicting real-world behaviors, such as how a spring oscillates while losing energy over time.

8. Summary Table

Here’s a concise table summarizing the key steps and formula for clarity:

Step	Description	Key Equation or Result
1. Set up integration by parts	Choose (u = e^{ax}), (dv = \cos(bx) , dx)	(\int u , dv = uv - \int v , du)
2. First application	Results in a new integral with sine	(I = \frac{e^{ax} \sin(bx)}{b} - \frac{a}{b} \int e^{ax} \sin(bx) , dx)
3. Second application	Apply integration by parts to the sine integral	Introduces back the original integral
4. Solve system	Combine equations and solve for (I)	(I = e^{ax} \frac{a \cos(bx) + b \sin(bx)}{a^2 + b^2} + C)
5. Final result	Compact formula with constant (C)	(\int e^{ax} \cos(bx) , dx = e^{ax} \frac{a \cos(bx) + b \sin(bx)}{a^2 + b^2} + C)

9. Summary and Key Takeaways

In summary, the integral (\int e^{ax} \cos(bx) , dx) is solved using integration by parts applied twice, leading to the formula (e^{ax} \frac{a \cos(bx) + b \sin(bx)}{a^2 + b^2} + C). This process involves careful algebraic manipulation and solving a system of equations, resulting in a clean expression that combines exponential and trigonometric terms.

Key Takeaways:

The solution depends on the constants (a) and (b), with (a^2 + b^2) in the denominator ensuring the formula is defined for most practical cases.
This integral is a great exercise in calculus, reinforcing techniques like integration by parts and the importance of verifying results through differentiation.
Real-world applications highlight its relevance, and practicing with numerical examples can solidify your understanding.

If you have more details, like specific values for (a) and (b), or if you’d like to explore related topics such as the sine version or definite integrals, feel free to ask—I’m here to help!

@Dersnotu