Answer
$\sec \theta -2e^{\theta}+C$
Work Step by Step
$\int (\sec \theta \tan \theta -2e^{\theta})d\theta=$$\int \sec\theta\tan\theta d\theta-2\int e^{\theta}d\theta=$$\sec \theta-2e^{\theta}+C$
Interval: $(n\pi-\pi/2,n\pi+\pi/2)$
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