Answer
$\frac{2}{7}(\cos\theta)^{7/2}-\frac{2}{3}(\cos\theta)^{3/2}+C$
Work Step by Step
$\int\sqrt{\cos\theta}\sin^3\theta\ d\theta$
$=\int\cos^{\frac{1}{2}}\theta\sin^2\theta\sin\theta\ d\theta$
$=\int\cos^{\frac{1}{2}}\theta(1-\cos^2\theta)\sin\theta\ d\theta$
Let $u=\cos\theta$. Then $du=-\sin\theta\ d\theta$, so $\sin\theta\ d\theta=-du$.
$=\int u^{\frac{1}{2}}(1-u^2)*(-1)du$
$=-\int(u^{\frac{1}{2}}-u^{\frac{5}{2}})du$
$=-\left(\frac{u^{\frac{3}{2}}}{\frac{3}{2}}-\frac{u^{\frac{7}{2}}}{\frac{7}{2}}+C\right)$
$=\frac{2u^{\frac{7}{2}}}{7}-\frac{2u^{\frac{3}{2}}}{3}+C$
$=\boxed{\frac{2}{7}(\cos\theta)^{7/2}-\frac{2}{3}(\cos\theta)^{3/2}+C}$
