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

$b x$

We have,

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

and $v=\frac{e^{a x}}{a^2+b^2}(a \cos b x+b \sin b x)$

∴  $\frac{u}{v}=\frac{a \sin b x-b \cos b x}{a \cos b x+b \sin b x}$

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

$\Rightarrow \frac{u}{v}=\tan (b x-\theta)$

$\Rightarrow \tan ^{-1} \frac{u}{v}=b x-\tan ^{-1} \frac{b}{a} \Rightarrow \tan ^{-1} \frac{u}{v}+\tan ^{-1} \frac{b}{a}=b x$