Chapter 5 - The Integral - 5.7 Substitution Method - Exercises - Page 276: 60

$$\frac{1}{6} (3 \sin x-1)^{2}+C$$

Given $$ \int \cos x(3 \sin x-1) d x $$ Let $$u=3 \sin x-1 \ \ \ \ \Rightarrow \ \ \ du =3\cos x dx $$ \begin{aligned} \int \cos x(3 \sin x-1) d x &=\frac{1}{3} \int u d u \\ &=\frac{1}{3}\left(\frac{1}{2} u^{2}+C\right) \\ &=\frac{1}{6} u^{2}+C \\ &= \frac{1}{6} (3 \sin x-1)^{2}+C \end{aligned}