Question

Find the integral by using the appropriate formula. $$ \int e^{2 x} \cos 3 x d x $$

Short answer

\(\int e^{2x} \cos(3x) \, dx = \frac{9}{13}(\frac{1}{3}e^{2x}\sin(3x) + \frac{2}{9}e^{2x}\cos(3x)) + C\)

Step-by-step solution

Assign u and dv

Assign \(u = e^{2x}\) and \(dv = \cos(3x) \, dx\) . Then find \(du\) and \(v\). So, \(du = 2e^{2x} \, dx\), and \(v = \frac{1}{3}\sin(3x)\).

Apply integration by parts

Apply the rule of integration by parts \(\int udv = uv - \int vdu\). This results in: \(\int e^{2x} \cos(3x) \, dx = uv - \int v\, du = e^{2x} * \frac{1}{3}\sin(3x) - \int \frac{1}{3}\sin(3x) * 2e^{2x} \, dx = \frac{1}{3}e^{2x}\sin(3x) - \frac{2}{3} \int e^{2x} \sin(3x) \, dx\)

Perform integration by parts again

We are left with another integral \(\int e^{2x} \sin(3x) \, dx\) . Let's perform integration by parts again, similarly assigining \(u = e^{2x}\) and \(dv = \sin(3x) \, dx\). Then, \(du = 2e^{2x} \, dx\) and \(v = -\frac{1}{3}\cos(3x)\). Apply the integration by parts formula results in: \(\int e^{2x} \sin(3x) \, dx = uv - \int v \, du = -\frac{1}{3}e^{2x}\cos(3x) - \int -\frac{1}{3}\cos(3x) * 2e^{2x} \, dx = -\frac{1}{3}e^{2x}\cos(3x) + \frac{2}{3} \int e^{2x} \cos(3x) \, dx\)

Finalize the Solution

Plug this result back into the integral we got from Step 2 : \(\frac{1}{3}e^{2x}\sin(3x) - \frac{2}{3}(-\frac{1}{3}e^{2x}\cos(3x) + \frac{2}{3} \int e^{2x} \cos(3x) \, dx) = \frac{1}{3}e^{2x}\sin(3x) + \frac{2}{9}e^{2x}\cos(3x) - \frac{4}{9} \int e^{2x} \cos(3x) \, dx\). Now, we see we have a similar integral on both sides of the equation. Let's move it all to the left side: \(\frac{13}{9}\int e^{2x} \cos(3x) \, dx = \frac{1}{3}e^{2x}\sin(3x) + \frac{2}{9}e^{2x}\cos(3x)\). Therefore, \(\int e^{2x} \cos(3x) \, dx = \frac{9}{13}(\frac{1}{3}e^{2x}\sin(3x) + \frac{2}{9}e^{2x}\cos(3x)) + C\), where \(C\) is the constant of integration.