Solving Integral of Exponential and Trigonometric Function
The image shows the integral:
\[
\int e^{2x} \cos(4x) \,dx
\]
To solve this integral, we'll have to use integration by parts multiple times or apply the tabular integration by parts method. However, there's an alternative approach that's often quicker and involves recognizing that the integral is the real part of a complex exponential. Here are the steps for the alternative method:
1. Recognize that \(e^{ix} = \cos(x) + i\sin(x)\), where \(i\) is the imaginary unit.
2. Write \(\cos(4x) = \Re(e^{4ix})\), where \(\Re\) denotes the real part:
\[
\int e^{2x} \cos(4x) \,dx = \Re\left(\int e^{2x} \cdot e^{4ix} \,dx\right)
\]
3. Combine the exponentials:
\[
\Re\left(\int e^{2x} \cdot e^{4ix} \,dx\right) = \Re\left(\int e^{(2+4i)x} \,dx\right)
\]
4. Integrate the complex exponential:
\[
\Re\left(\int e^{(2+4i)x} \,dx\right) = \Re\left(\frac{1}{2+4i} e^{(2+4i)x} + C\right)
\]
where \(C\) is the constant of integration.
5. Simplify the expression by finding the real part after multiplying by the conjugate to get rid of the \(i\):
\[
\frac{1}{2+4i} = \frac{2-4i}{(2+4i)(2-4i)} = \frac{2-4i}{4+16} = \frac{2-4i}{20} = \frac{1}{10} - \frac{1}{5}i
\]
6. Insert this into the integrated function:
\[
\Re\left(\frac{1}{10} - \frac{1}{5}i\right)e^{(2+4i)x} + C = \left(\frac{1}{10} e^{2x} \cos(4x) - \frac{1}{5} e^{2x} \sin(4x)\right) + C
\]
7. The final result only requires the real part:
\[
\int e^{2x} \cos(4x) \,dx = \frac{1}{10} e^{2x} \cos(4x) - \frac{1}{5} e^{2x} \sin(4x) + C
\]
Hence, this is the evaluated integral of the original function.
