integrating by parts $9e^{2x}\cos(3x)$

Question

integrating by parts:

$$\int 9e^{2x}\cos(3x)dx$$

It seems like whichever part I start with integrating or deriving It still leads to a "by parts" integral. How Do I deal with it?

Answer 1

Here’s a nice and easy way to evaluate the integral using the tabular method. First, choose a function to differentiate. We’ll let our $u$ be $\cos 3x$ and $\mathrm dv$ to be $e^{3x}$. Setting up the table gives$$\begin{array}{|c|c|c|}\hline \text{sign} & u & \mathrm dv\\\hline+ & \cos 3x & e^{3x}\\\hline\end{array}$$Now differentiate and integrate $u$ and $\mathrm dv$ respectively until you either return back to your original integral or when $u$ reaches zero (when you differentiate a constant).$$\begin{array}{|c|c|c|}\hline\text{sign} & u & \mathrm dv\\\hline + & \cos 3x & e^{3x}\\\hline - & -3\sin 3x & \frac 13e^{3x}\\\hline + & -9\cos 3x & \frac 19e^{3x}\\\hline\end{array}$$Hence, if we call the integral $\mathfrak{I}$, then$$\begin{align*}\mathfrak{I} & =\frac 13e^{3x}\cos 3x+\frac 13e^{3x}\sin 3x-\int\mathrm dx\, e^{3x}\cos 3x\\ & =\frac 13e^{3x}\cos 3x+\frac 13e^{3x}\sin 3x-\mathfrak{I}\end{align*}$$Therefore, isolating $\mathfrak{I}$, we get that$$\int\mathrm dx\, e^{3x}\cos 3x\color{blue}{=\frac 16e^{3x}\cos 3x+\frac 16e^{3x}\sin 3x+C}$$

Answer 2

Straightforward... use:

$$\int f(x) g'(x)\ dx = f g - \int f'(x) g(x)\ dx$$

where $g \propto e^{2 x}$ and $f \propto \cos (3 x)$.

Result:

$$\frac{9}{13} e^{2 x} (3 \sin (3 x)+2 \cos (3 x))$$

Answer 3

If you need $$I=\int e^{ax}\cos(bx)\,dx$$ consider instead $$J=\int e^{ax} e^{ibx}\,dx=\int e^{(a+i b)x}\,dx=\frac{e^{(a+i b)x} }{a+ib }$$ and take the real part of it.

No need of integration by parts.