# Where does this 1 come from when balancing this integral equation?

$$\int e^{ax}\cos(bx)\,\mathrm dx = \frac1{a}e^{ax}\cos(bx) + \frac{b}{a^2}e^{ax}\sin(bx) - \frac{b^2}{a^2}\int e^{ax}\cos(bx)\,\mathrm dx$$

$$\left(1 + \frac{b^2}{a^2}\right)\int e^{ax}\cos(bx)\,\mathrm dx = \frac{1}{a}e^{ax}\cos(bx) + \frac{b}{a^2}e^{ax}\sin(bx) + C$$

Where does the $1$ in $\displaystyle \left(1 + \frac{b^2}{a^2}\right)$ come from?

-

Yeah so they have taken the quantity $-b^{2}/a^{2}$ to the Left hand side. So you have the LHS as $$1 \cdot \int e^{ax}\cos(bx) \rm{dx} + \frac{b^{2}}{a^{2}} \int e^{ax}\cos(bx) \ \text{dx} = \Bigl(1+\frac{b^{2}}{a^{2}}\Bigr)\int e^{ax}\cos(bx) \ \text{dx}$$

-
If $a \neq 0$ and you multiply $a$ with $1$ you get $a$ itself. –  user9413 May 16 '11 at 8:53
I guess there is a 1 there. Thank you =) –  Louis May 16 '11 at 9:00