1. We are asked to find the integral of the function $f(x) = x \cdot \sin(x^2)$. 2. The integral we want to compute is $$\int x \sin(x^2) \, dx.$$ 3. Use substitution: let $u = x^2$, then $du = 2x \, dx$ or equivalently $x \, dx = \frac{du}{2}$. 4. Substitute into the integral: $$\int x \sin(x^2) \, dx = \int \sin(u) \cdot \frac{du}{2} = \frac{1}{2} \int \sin(u) \, du.$$ 5. The integral of $\sin(u)$ is $-\cos(u)$, so $$\frac{1}{2} \int \sin(u) \, du = \frac{1}{2} (-\cos(u)) + C = -\frac{1}{2} \cos(u) + C.$$ 6. Substitute back $u = x^2$: $$\int x \sin(x^2) \, dx = -\frac{1}{2} \cos(x^2) + C.$$ **Final answer:** $$\boxed{\int x \sin(x^2) \, dx = -\frac{1}{2} \cos(x^2) + C}.$$