Chapter 5: Integrals - Section 5.5 - Indefinite Integrals and the Substitution Method - Exercises 5.5 - Page 295: 8

$\int x\sin (2x^2) dx = -\frac{1}{4}\cos(2x^2) + C$

$\int x\sin (2x^2) dx\space$ and $\space u = 2x^2$ $du = 4x\cdot dx$ Doing the substitution $u = 2x^2$ $\int \sin (u) \frac{du}{4}\space \Rightarrow\space \frac{1}{4}\int \sin(u)du$ Applying the integrative rules $\frac{1}{4}\int \sin(u)du\space = -\frac{1}{4}\cos(u) + C$ Backing to $x$ $\int x\sin (2x^2) dx = -\frac{1}{4}\cos(2x^2) + C$