Chapter 8 - Techniques of Integration - 8.2 Trigonometric Integrals - Exercises - Page 403: 5

$$-\frac{1}{4} \cos ^{4} t+\frac{1}{6} \cos ^{6} t+c$$

\begin{aligned} \int \sin ^{3} t \cos ^{3} t d t &=\int \sin ^{2} t \sin t \cos ^{3} t d t \\ &=\int\left(1-\cos ^{2} t\right) \sin t \cos ^{3} t d t \\ &=\int\left(\cos ^{3} t-\cos ^{5} t\right) \sin t d t \\ &=-\frac{1}{4} \cos ^{4} t+\frac{1}{6} \cos ^{6} t+c \end{aligned}