To integrate $e^{2x} \sin(4x)$, we can use the method of integration by parts in conjunction with differentiation under the integral sign or by using a systematic approach of solving a second-order linear differential equation.

Here's how we proceed:

We need to solve the integral: $I = \int e^{2x} \sin(4x) \, dx$

Step 1: Integration by parts

The trick is to apply integration by parts twice. Let: $I = \int e^{2x} \sin(4x) \, dx$

Use the tabular method (which is a shortcut for integration by parts). For the first part:

1. Differentiate $e^{2x}$ repeatedly.
2. Integrate $\sin(4x)$ repeatedly.

First integration by parts:

Let:

• $u = e^{2x}$ (differentiate this),
• $dv = \sin(4x) dx$ (integrate this).

Now differentiate and integrate as follows:

• $u = e^{2x} \quad \Rightarrow \quad du = 2e^{2x} dx$
• $dv = \sin(4x) dx \quad \Rightarrow \quad v = -\frac{1}{4} \cos(4x)$

Now apply integration by parts: $I = u v - \int v \, du$ $I = e^{2x} \left( -\frac{1}{4} \cos(4x) \right) - \int \left( -\frac{1}{4} \cos(4x) \right) (2e^{2x}) dx$ $I = -\frac{1}{4} e^{2x} \cos(4x) + \frac{1}{2} \int e^{2x} \cos(4x) dx$

Now we need to solve the integral $\int e^{2x} \cos(4x) \, dx$.

Second integration by parts:

Let’s call this integral $J = \int e^{2x} \cos(4x) \, dx$. We use integration by parts again:

• $u = e^{2x} \quad \Rightarrow \quad du = 2e^{2x} dx$
• $dv = \cos(4x) dx \quad \Rightarrow \quad v = \frac{1}{4} \sin(4x)$

Now apply integration by parts again: $J = u v - \int v \, du$ $J = e^{2x} \left( \frac{1}{4} \sin(4x) \right) - \int \left( \frac{1}{4} \sin(4x) \right) (2e^{2x}) dx$ $J = \frac{1}{4} e^{2x} \sin(4x) - \frac{1}{2} \int e^{2x} \sin(4x) dx$

Notice that the integral on the right is the original integral $I$, so we can substitute this back: $J = \frac{1}{4} e^{2x} \sin(4x) - \frac{1}{2} I$

Step 2: Solve for $I$

Now substitute this result for $J$ back into the expression for $I$: $I = -\frac{1}{4} e^{2x} \cos(4x) + \frac{1}{2} J$ $I = -\frac{1}{4} e^{2x} \cos(4x) + \frac{1}{2} \left( \frac{1}{4} e^{2x} \sin(4x) - \frac{1}{2} I \right)$ $I = -\frac{1}{4} e^{2x} \cos(4x) + \frac{1}{8} e^{2x} \sin(4x) - \frac{1}{4} I$

Now collect terms involving $I$: $I + \frac{1}{4} I = -\frac{1}{4} e^{2x} \cos(4x) + \frac{1}{8} e^{2x} \sin(4x)$ $\frac{5}{4} I = -\frac{1}{4} e^{2x} \cos(4x) + \frac{1}{8} e^{2x} \sin(4x)$

Finally, solve for $I$: $I = \frac{1}{5} \left( -e^{2x} \cos(4x) + \frac{1}{2} e^{2x} \sin(4x) \right)$

Final Answer:

$\int e^{2x} \sin(4x) \, dx = \frac{1}{5} e^{2x} \left( -\cos(4x) + \frac{1}{2} \sin(4x) \right) + C$

Would you like any further clarification or details?

Here are some related questions you might want to explore:

1. How do you apply integration by parts in different cases?
2. How do you solve second-order differential equations related to exponential functions?
3. Can we generalize this method for other integrals of exponential and trigonometric functions?
4. What is the role of Euler's formula in simplifying such integrals?
5. How can Laplace transforms be used to solve such integrals?

Tip: When integrating products of exponential and trigonometric functions, integrating by parts twice often leads to a solvable system of equations.