Step 1: Choose the substitution t = sqrt(x).
The integrand involves $\sqrt{x}$, so let $t = \sqrt{x}$, giving $x = t^2$ and $dx = 2t\,dt$. New limits: $x=0 \Rightarrow t=0$; $x=\frac{\pi^2}{4} \Rightarrow t=\frac{\pi}{2}$.
Step 2: Transform the integral.
Replacing $\sqrt{x}$ with $t$ and $dx$ with $2t\,dt$: \[\int_0^{\pi^2/4}(2\sin\sqrt{x} + \sqrt{x}\cos\sqrt{x})\,dx = \int_0^{\pi/2}(2\sin t + t\cos t) \cdot 2t\,dt = \int_0^{\pi/2}(4t\sin t + 2t^2\cos t)\,dt\]
Step 3: Recognize the integrand as an exact derivative.
Notice that $\frac{d}{dt}(2t^2\sin t) = 4t\sin t + 2t^2\cos t$ (by the product rule). This matches the integrand exactly -- a key insight!
Step 4: Apply the fundamental theorem of calculus.
\[\int_0^{\pi/2}(4t\sin t + 2t^2\cos t)\,dt = \left[2t^2\sin t\right]_0^{\pi/2}\]
Step 5: Evaluate at the limits.
At $t = \frac{\pi}{2}$: $2\!\left(\frac{\pi}{2}\right)^{\!2}\!\sin\frac{\pi}{2} = 2 \cdot \frac{\pi^2}{4} \cdot 1 = \frac{\pi^2}{2}$. At $t = 0$: $2(0)^2\sin 0 = 0$. So the result is $\frac{\pi^2}{2} - 0 = \frac{\pi^2}{2}$.
Step 6: State the final answer.
\[\boxed{\dfrac{\pi^2}{2}}\]