Chapter 5 - The Integral - 5.7 Substitution Method - Exercises - Page 275: 48

$$-\frac{1}{2} \cos \theta^{2}+C $$

Given $$ \int \theta \sin \left(\theta^{2}\right) d \theta $$ Let $$u=\theta^{2}\ \ \ \Rightarrow \ \ \ du=2\theta d \theta $$ Then \begin{aligned} \int \theta \sin \left(\theta^{2}\right) d \theta &=\frac{1}{2} \int \sin u d u \\ &=\frac{1}{2}(-\cos u+C) \\ &=-\frac{1}{2} \cos u+C \\ &=-\frac{1}{2} \cos \theta^{2}+C \end{aligned}