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$ \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?

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

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    I guess there is$a$1 there. Thank you =)2011-05-16