Chapter 5: Integrals - Section 5.5 - Indefinite Integrals and the Substitution Method - Exercises 5.5 - Page 295: 19

$\frac{-2}{5}(1-\theta^2)^{5/4} +C$

Let $u=1-\theta^2=>du=-2\theta d\theta=>\frac{-1}{2}du=\theta d\theta$ =$\int \theta \sqrt[4] {1-\theta^2}d \theta$ =$\int \sqrt[1]{u}(-\frac{1}{2}du)$ =$(\frac{-1}{2})(\frac{4}{5}u^{5/4})+C$ =$\frac{-2}{5}(1-\theta^2)^{5/4} +C$