Step 1: Spot the structure of the integrand.
The integrand $x\sin^2(x^2)$ pairs a function of $x^2$ with a single factor of $x$. That single $x$ is the tell-tale sign that the substitution $t = x^2$ will collapse things neatly.
Step 2: Carry out the substitution.
Let $t = x^2$, so $dt = 2x\,dx$, giving $x\,dx = \dfrac{dt}{2}$. The limits map as $x = 0 \Rightarrow t = 0$ and $x = \sqrt{\pi} \Rightarrow t = \pi$. Hence \[ I = \frac{1}{2}\int_{0}^{\pi} \sin^2 t \, dt \]
Step 3: Bring in a symmetry shortcut for $\sin^2$.
Over a full half-period from $0$ to $\pi$, the average value of $\sin^2 t$ is exactly $\tfrac{1}{2}$, because $\sin^2 t$ and $\cos^2 t$ are mirror images that add to 1 and share the same average.
Step 4: Use the average to evaluate the inner integral.
Average value $\tfrac12$ over an interval of length $\pi$ gives \[ \int_{0}^{\pi}\sin^2 t\,dt = \frac{1}{2}\times \pi = \frac{\pi}{2} \]
Step 5: Reinsert the front factor.
\[ I = \frac{1}{2}\times \frac{\pi}{2} = \frac{\pi}{4} \]
Step 6: Sanity-check against direct integration.
Using $\sin^2 t = \dfrac{1 - \cos 2t}{2}$ would give $\dfrac14\big[t - \tfrac{\sin 2t}{2}\big]_0^\pi = \dfrac{\pi}{4}$, the same result. So the average-value shortcut agrees with the key.
\[ \boxed{I = \frac{\pi}{4}} \]