Question

Show that $\int e^{-ax}\cos bxdx=\frac{e^{-ax}}{r}\cos (bx+\alpha ).$ where $r=\sqrt{a^2+b^2}$ and $\alpha ={\tan }^{-1}\left(b/a\right).$

Answer 1

Let $I=\int e^{-ax}\cos bxdx$
Using $\int e^{px}\cos qxdx=\frac{e^{px}}{p^2+q^2}(p\cos qx+q\sin qx)+c$
$I=\frac{e^{-ax}}{a^2+b^2}(-a\cos bx+b\sin bx)$
$=\frac{e^{-ax}}{a^2+b^2}(\cos bx+{\tan }^{-1}\left(\frac{b}{a}\right))$
$\therefore r=\frac{1}{a^2+b^2}$ and $\alpha ={\tan }^{-1}\frac{b}{a}$