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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?

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    $\begingroup$ Do the "by parts" twice, you get back to original and you can just solve the equation. $\endgroup$
    – orion
    Commented Jan 18, 2019 at 20:23
  • $\begingroup$ Integrate by parts twice, & then do some linear algebra. $\endgroup$ Commented Jan 18, 2019 at 20:23

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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$$\int\mathrm dx\, e^{3x}\cos 3x\color{blue}{=\frac 16e^{3x}\cos 3x+\frac 16e^{3x}\sin 3x+C}$$

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  • $\begingroup$ Yup!!!! Thankss $\endgroup$
    – NPLS
    Commented Jan 18, 2019 at 23:21
  • $\begingroup$ @NPLS I just realized, your problem is $e^{\color{red}{2}x}$ instead of $e^{\color{blue}{3}x}$. Anyways, I hope you get the point of my answer. $\endgroup$
    – Frank W
    Commented Jan 19, 2019 at 18:38
  • $\begingroup$ yeah yeah I got it $\endgroup$
    – NPLS
    Commented Jan 19, 2019 at 23:22
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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:

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

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  • $\begingroup$ I'd like the explanation, on how to derive the resut $\endgroup$
    – NPLS
    Commented Jan 18, 2019 at 20:24
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    $\begingroup$ @NPLS : Integrate by parts twice. You end up with something like $\int = f + c\cdot\int$, so you conclude that $\int = f/(1-c)$. $\endgroup$
    – MPW
    Commented Jan 18, 2019 at 20:28
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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.

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