Chapter 5 - The Integral - 5.7 Substitution Method - Exercises - Page 276: 52

$$\frac{1}{9} (\sin \theta)^{9}+C$$

Given $$\int \sin ^{8} \theta \cos \theta d \theta$$ Let $$ u=\sin \theta \ \ \ \ \Rightarrow \ \ \ du = \cos \theta d \theta$$ Then \begin{aligned} \int \sin ^{8} \theta \cos \theta d \theta &=\int u^{8} d u \\ &=\frac{1}{9} u^{9}+C\\ &=\frac{1}{9} (\sin \theta)^{9}+C \end{aligned}