## Calculus 10th Edition

$-\frac{ln|cos(5θ)|}{5}+C$
$\int tan (5θ) dθ$ let $u=5θ$ $du= 5θ$ $dθ=\frac{1}{5}du$ $\int tan (5θ) dθ$ $=\frac{1}{5}\int tan (u)$ $du$ $=\frac{1}{5}(-ln|cos(u)|)+C$ $=-\frac{ln|cos(5θ)|}{5}+C$