## Calculus 10th Edition

$\frac{1}{3}sin(3θ)-θ+C$
$\int cos(3θ -1) dθ$ $=\int cos(3θ) dθ$ $-\int 1$ $dθ$ for $\int cos(3θ) dθ$, let $u=3θ$ $du=3$ $dθ$ $dθ=\frac{1}{3}du$ $\int cos(3θ) dθ$ $=\int cos (u)$ $\frac{1}{3}du$ $=\frac{1}{3}\int cos(u)$ $du$ $=\frac{1}{3}(sin(u))$ $=\frac{1}{3}sin(3θ)$ $=\int cos(3θ) dθ$ $-\int 1$ $dθ$ $=\frac{1}{3}sin(3θ)-θ+C$