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It's the part before and after "Thus".

Pic

$$I = \ldots = \int e^{ax} \cos bx \ \mathrm{d}x = \frac{1}{b} e^{ax} \sin bx + \frac{a}{b^{2}} \cos bx - \frac{a^{2}}{b^{2}} I.$$
Thus
$$\left( 1 + \frac{a^{2}}{b^{2}} \right) I = \frac{1}{b} e^{ax} \sin bx + \frac{a}{b^{2}} \cos bx + C_1,$$
and
$$\int e^{ax} \cos bx \ \mathrm{d}x = I = \frac{be^{ax}\sin bx+ae^{ax} \cos bx}{b^2+a^2}+C.$$

Were does the "+1" come from? I thought this was an old "move to other side of equal-sign" until that +1 spawned in my face hehe.

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  • $\begingroup$ Ummmm......what? $\endgroup$
    – Aryabhata
    Commented Apr 12, 2011 at 17:31
  • $\begingroup$ What is $C_1$ after thus? $\endgroup$
    – Fabian
    Commented Apr 12, 2011 at 17:33
  • $\begingroup$ I'll update with pic entire example. Two secs. $\endgroup$
    – Algific
    Commented Apr 12, 2011 at 17:37

1 Answer 1

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The integral calculation is done by integration by parts: \begin{align} I = \int e^{ax} \cos bx \ dx = \frac{1}{b} e^{ax} \sin bx + \frac{a}{b^{2}} \cos bx - \frac{a^{2}}{b^{2}} I. \end{align} Solving for $I$, we have \begin{align} I + \frac{a^{2}}{b^{2}} I = \left( 1 + \frac{a^{2}}{b^{2}} \right) I = \frac{1}{b} e^{ax} \sin bx + \frac{a}{b^{2}} \cos bx + C_1, \end{align} (answering your $+1$ question) so \begin{align} I = \left( 1 + \frac{a^2}{b^2} \right)^{-1} \left( \frac{1}{b} e^{ax} \sin bx + \frac{a}{b^{2}} \cos bx \right) + C = \frac{b e^{a x} \sin bx + a e^{ax} \cos bx }{a^{2} + b^{2}} + C \end{align} by multiplying by $\frac{b^{2}}{b^{2}}$ and clearing the denominator, and where we have added $C_1$ and $C$ as arbitrary integration constants, related by the factor $(1 + \frac{a^2}{b^2})^{-1}$.

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    $\begingroup$ Really wish they could put this in the book for beginners, it would cost so little. From a pedagogy perspective I find most mathbooks to be a bitch. It's like you are suppose to be punished for first starting to take education seriously at an adult level. $\endgroup$
    – Algific
    Commented Apr 12, 2011 at 17:46
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    $\begingroup$ I felt like that when I was going through Baby Rudin. $\endgroup$
    – Michael Chen
    Commented Apr 12, 2011 at 18:27

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