If $u=\int e^{ax}\sin bxdx\text{ and }v=\int e^{ax}\cos bxdx,$ then ${\tan }^{-1}\left(\frac{u}{v}\right)+{\tan }^{-1}\left(\frac{b}{a}\right)$ equals

We have,

       $u=\frac{e^{ax}}{a^2+b^2}(a\sin bx-b\cos bx)$

and,  $v=-\frac{e^{ax}}{a^2+b^2}(a\cos bx+b\sin bx)$

$\therefore \frac{u}{v}=\frac{a\sin bx-b\cos bx}{a\cos bx+b\sin bx}$

$\Rightarrow \frac{u}{v}=\frac{\sin (bx-\theta )}{\cos (bx-\theta )},$ where $a=r\cos \theta ,b=r\sin \theta$ and $\tan \theta =\frac{b}{a}$

$\begin{array}{l}\Rightarrow \frac{u}{v}=\tan (bx-0)\\ \Rightarrow {\tan }^{-1}\frac{u}{v}=bx-{\tan }^{-1}\frac{b}{a}\Rightarrow {\tan }^{-1}\frac{u}{v}+{\tan }^{-1}\frac{b}{a}=bx\end{array}$