Answer
$3 \sec (\dfrac{\theta}{3})+C$
Work Step by Step
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} $
$=3 \sec (\dfrac{\theta}{3})+C$
