Answer
$$\int \sin^22\theta\cos^32\theta d\theta=-\frac{\sin2\theta\cos^42\theta}{10}+\frac{\sin2\theta\cos^22\theta}{30}+\frac{\sin2\theta}{15}+C$$
Work Step by Step
$$A=\int \sin^22\theta\cos^32\theta d\theta$$
Use Reduction Formula 75 to reduce $\sin^22\theta$ first.
$$\int\sin^nax\cos^maxdx=-\frac{\sin^{n-1}ax\cos^{m+1}ax}{a(m+n)}+\frac{n-1}{m+n}\int\sin^{n-2}ax\cos^maxdx$$
for $a=2$, $m=3$ and $n=2$
$$A=-\frac{\sin2\theta\cos^42\theta}{10}+\frac{1}{5}\int\cos^32\theta d\theta$$
Use Reduction Formula 76 to reduce $\cos^32\theta$.
$$\int\sin^nax\cos^maxdx=\frac{\sin^{n+1}ax\cos^{m-1}ax}{a(m+n)}+\frac{m-1}{m+n}\int\sin^{n}ax\cos^{m-2}axdx$$
for $a=2$, $m=3$ and $n=0$
$$A=-\frac{\sin2\theta\cos^42\theta}{10}+\frac{1}{5}\Big(\frac{\sin2\theta\cos^22\theta}{6}+\frac{2}{3}\int\cos2\theta d\theta\Big)$$ $$A=-\frac{\sin2\theta\cos^42\theta}{10}+\frac{1}{5}\Big(\frac{\sin2\theta\cos^22\theta}{6}+\frac{2}{3}\times\frac{\sin2\theta}{2}\Big)+C$$ $$$A=-\frac{\sin2\theta\cos^42\theta}{10}+\frac{1}{5}\Big(\frac{\sin2\theta\cos^22\theta}{6}+\frac{\sin2\theta}{3}\Big)+C$$ $$A=-\frac{\sin2\theta\cos^42\theta}{10}+\frac{\sin2\theta\cos^22\theta}{30}+\frac{\sin2\theta}{15}+C$$
