Step 1: Rewrite each square root using half-angle-style algebra without immediately combining:
$\sqrt{\cot x}=\sqrt{\cos x}/\sqrt{\sin x}$ and $\sqrt{\tan x}=\sqrt{\sin x}/\sqrt{\cos x}$; putting both over the common denominator $\sqrt{\sin x\cos x}$ gives $\dfrac{\cos x+\sin x}{\sqrt{\sin x\cos x}}$, same as before but arrived at by treating each root as a separate fraction first.
Step 2: Instead of substituting $t=\sin x-\cos x$, verify the alternate substitution $u=\sin x+\cos x$ does NOT simplify as cleanly, to justify why the difference form is the right choice:
$du=(\cos x-\sin x)dx$, which does not match the $(\sin x+\cos x)$ appearing in the numerator — confirming $t=\sin x-\cos x$ (matching the numerator via its derivative) is the correct substitution.
Step 3: Proceed with $t=\sin x-\cos x$, $dt=(\sin x+\cos x)dx$, and rewrite $\sin2x$ in terms of $t$:
$t^2=1-\sin2x\Rightarrow\sin2x=1-t^2$, giving $\displaystyle\int\dfrac{\sqrt2\,dt}{\sqrt{1-t^2}}$.
Step 4: Apply the standard arcsine antiderivative:
$\sqrt2\sin^{-1}(t)+C$.
Final Answer:
\[ \boxed{\sqrt2\,\sin^{-1}(\sin x-\cos x)+C} \]