Answer
$$\int\sqrt{\cos\theta}\sin^3\theta d\theta=\frac{2(\cos\theta)^{7/2}}{7}-\frac{2(\cos\theta)^{3/2}}{3}+C$$
Work Step by Step
$$A=\int\sqrt{\cos\theta}\sin^3\theta d\theta$$
$$A=\int(\cos\theta)^{1/2}\sin^3\theta d\theta$$
Since the factor of $\sin$ is odd, we would change $\sin$ in terms of $\cos$ and set aside 1 factor of $\sin$ for later substitution.
$$A=\int(\cos\theta)^{1/2}\sin^2\theta\sin\theta d\theta$$
$$A=\int(\cos\theta)^{1/2}(1-\cos^2\theta)\sin\theta d\theta$$
Now let $u=\cos\theta$
We would have $du=-\sin\theta d\theta$, or $\sin\theta d\theta=-du$
That means $$A=-\int u^{1/2}(1-u^2) du$$
$$A=-\int (u^{1/2}-u^{5/2}) du$$
$$A=\int (u^{5/2}-u^{1/2}) du$$
$$A=\Bigg[\frac{u^{7/2}}{7/2}-\frac{u^{3/2}}{3/2}\Bigg]+C$$
$$A=\frac{2u^{7/2}}{7}-\frac{2u^{3/2}}{3}+C$$
$$A=\frac{2(\cos\theta)^{7/2}}{7}-\frac{2(\cos\theta)^{3/2}}{3}+C$$
