 prove that integration of e^ax cosbxdx = e^ax/a^2+b^2 (acosbx+bsindx)+c 

We are tasked with proving that:

\[
\int e^{ax} \cos(bx) \, dx = \frac{e^{ax}}{a^2 + b^2} \left( a \cos(bx) + b \sin(bx) \right) + C
\]

Let's go step by step to derive this.

### Step 1: Use Integration by Parts
We apply **integration by parts**, using the formula:

\[
\int u \, dv = uv - \int v \, du
\]

Let:
- \( u = e^{ax} \) so that \( du = a e^{ax} dx \)
- \( dv = \cos(bx) dx \) so that \( v = \frac{\sin(bx)}{b} \)

Now, apply the integration by parts formula:

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

This simplifies to:

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

### Step 2: Solve \( \int e^{ax} \sin(bx) \, dx \)
We now need to compute the integral \( \int e^{ax} \sin(bx) \, dx \). We again use integration by parts.

Let:
- \( u = e^{ax} \) so that \( du = a e^{ax} dx \)
- \( dv = \sin(bx) dx \) so that \( v = -\frac{\cos(bx)}{b} \)

Now, apply the integration by parts formula again:

\[
\int e^{ax} \sin(bx) \, dx = -\frac{e^{ax} \cos(bx)}{b} + \frac{a}{b} \int e^{ax} \cos(bx) \, dx
\]

### Step 3: Set up a system of equations
We now have the following system of equations:
1. \( I = \frac{e^{ax} \sin(bx)}{b} - \frac{a}{b} J \)
2. \( J = -\frac{e^{ax} \cos(bx)}{b} + \frac{a}{b} I \)

Where:
- \( I = \int e^{ax} \cos(bx) \, dx \)
- \( J = \int e^{ax} \sin(bx) \, dx \)

Substitute the expression for \( J \) into the first equation for \( I \):

\[
I = \frac{e^{ax} \sin(bx)}{b} - \frac{a}{b} \left( -\frac{e^{ax} \cos(bx)}{b} + \frac{a}{b} I \right)
\]

Simplify the right-hand side:

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

### Step 4: Solve for \( I \)
Now, collect all terms involving \( I \) on the left-hand side:

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

Factor \( I \) out of the left-hand side:

\[
I \left( 1 + \frac{a^2}{b^2} \right) = \frac{e^{ax}}{b^2} \left( a \cos(bx) + b \sin(bx) \right)
\]

Finally, solve for \( I \):

\[
I = \frac{e^{ax}}{a^2 + b^2} \left( a \cos(bx) + b \sin(bx) \right)
\]

### Step 5: Add the constant of integration
Don't forget to add the constant of integration, \( C \), giving us the final result:

\[
\int e^{ax} \cos(bx) \, dx = \frac{e^{ax}}{a^2 + b^2} \left( a \cos(bx) + b \sin(bx) \right) + C
\]

This completes the proof.

---

Would you like a more detailed explanation on any step or have questions? Here are some related questions to explore further:

1. How does integration by parts work, and when is it useful?
2. Can you apply a similar method to \( \int e^{ax} \sin(bx) \, dx \)?
3. Why do we need to solve the system of equations involving \( I \) and \( J \)?
4. What are some practical applications of integrals like this one in physics or engineering?
5. How can you verify the result of this integration using differentiation?

**Tip**: When solving integration problems involving products of exponentials and trigonometric functions, integration by parts is often the most effective approach.

Prove that integration of e^ax cos(bx) dx = (e^ax / (a^2 + b^2)) (a cos(bx) + b sin(bx)) + C

Learn how to prove the integral of e^ax * cos(bx) dx with a detailed step-by-step solution. This proof involves integration by parts, solving a system of equations, and using exponential and trigonometric functions. Ideal for advanced high school or undergraduate students.

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