Answer
$\tan{\theta}+\cos{\theta}+C$
Work Step by Step
$\int(\sec^2{\theta}-\sin{\theta})d\theta=\int\sec^2{\theta}d\theta+\int(-\sin\theta)d\theta$
$=\tan{\theta}+\cos{\theta}+C$
CHECK:
$\frac{d}{d\theta}(\tan{\theta}+\cos{\theta}+C)$
$\frac{d}{d\theta}\tan{\theta}+\frac{d}{d\theta}\cos{\theta}+\frac{d}{d\theta}C=\sec^2{\theta}+-\sin{\theta}+0=\sec^2{\theta}-\sin{\theta}$
