## University Calculus: Early Transcendentals (3rd Edition)

$3 \sec (\dfrac{\theta}{3})+C$
Calculate the anti-derivative: Since, we know $\int \sec x \tan x =\sec x+C$ Substitute $k=\dfrac{\theta}{3} \implies dk=(1/3) d \theta$ $=3 \int \sec k \tan k+C$ or, $=3 \sec k+C$ Back substitution: $k=\dfrac{\theta}{3}$ $=3 \sec (\dfrac{\theta}{3})+C$