# Another magic re-write question, picture!

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

$$I = \ldots = \int e^{ax} \cos bx \ \mathrm{d}x = \frac{1}{b} e^{ax} \sin bx + \frac{a}{b^{2}} \cos bx - \frac{a^{2}}{b^{2}} I.$$
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.

-
Ummmm......what? – Aryabhata Apr 12 '11 at 17:31
What is $C_1$ after thus? – Fabian Apr 12 '11 at 17:33
I'll update with pic entire example. Two secs. – Algific Apr 12 '11 at 17:37

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 $(1 + \frac{a^2}{b^2})^{-1}$.