Chapter 5 - Section 5.5 - Indefinite Integrals and the Substitution Method - Exercises - Page 329: 8

$-\frac{1}{4}(cos(2x^2))+c$

$u=2x^2$, therefore $du=4xdx$. Let's rewrite the integral! $\int (xsin(2x^2))dx$ We do the substitution! $\int (xsin(2x^2))dx=\int (sinu)\frac{du}{4}=\frac{1}{4}\int (sinu)du=\frac{1}{4}(-cosu)+c$ We re-do the substitution! $-\frac{1}{4}(cosu)+c=-\frac{1}{4}(cos(2x^2))+c$