Answer
$$\int\theta\sqrt[4]{1-\theta^2}d\theta=-\frac{2(1-\theta^2)^{5/4}}{5}+C$$
Work Step by Step
$$A=\int\theta\sqrt[4]{1-\theta^2}d\theta$$
We set $u=1-\theta^2$.
Then $$du=-2\theta d\theta$$
That means, $$\theta d\theta=-\frac{1}{2}du$$
Therefore, $$A=-\frac{1}{2}\int\sqrt[4]udu=-\frac{1}{2}\int u^{1/4}du$$ $$A=-\frac{1}{2}\times\frac{u^{5/4}}{\frac{5}{4}}+C=-\frac{u^{5/4}}{\frac{5}{2}}+C=-\frac{2u^{5/4}}{5}+C$$ $$A=-\frac{2(1-\theta^2)^{5/4}}{5}+C$$
