Step 1: Dividing numerator and denominator by cos x first:
\(\dfrac{\sin x+\cos x}{\sqrt{\sin x\cos x}}\): multiply numerator and denominator inside the root by 2 to prepare for the double-angle identity: \(\sin x\cos x=\dfrac{\sin2x}{2}\), so the integrand is \(\dfrac{\sin x+\cos x}{\sqrt{\sin2x/2}}=\sqrt2\cdot\dfrac{\sin x+\cos x}{\sqrt{\sin2x}}\).
Step 2: Relating sin2x to (sinx − cosx):
Since \((\sin x-\cos x)^2=1-\sin2x\), write \(\sin2x=1-(\sin x-\cos x)^2\).
Step 3: Substituting t = sinx − cosx:
With \(dt=(\cos x+\sin x)dx\) and \(\sin2x=1-t^2\): the integral becomes \(\sqrt2\displaystyle\int\dfrac{dt}{\sqrt{1-t^2}}\), the same standard arcsin form reached via the other route.
Step 4: Integrating:
\(\sqrt2\sin^{-1}t+C\).
Step 5: Back-substituting:
\(t=\sin x-\cos x\), confirming the identical final antiderivative.
Final Answer:
\[ \boxed{\sqrt2\,\sin^{-1}(\sin x-\cos x)+C} \]