Answer
$\tan(\theta)+\sec(\theta)+C$
Work Step by Step
Multiply the integrand by $\frac{1+\sin(\theta)}{1+\sin(\theta)}$ to obtain $\displaystyle\int\frac{1+\sin(\theta)}{1-\sin^2(\theta)}d\theta=\int\frac{1+\sin(\theta)}{\cos^2(\theta)}d\theta=\int\sec^2(\theta)+\sec(\theta)\tan(\theta)d\theta$. Recall that $\frac{d}{d\theta}\tan(\theta)=\sec^2(\theta)$ and that $\frac{d}{d\theta}\sec(\theta)=\sec(\theta)\tan(\theta)$. Therefore:
$\displaystyle\int\sec^2(\theta)+\sec(\theta)\tan(\theta)d\theta=\fbox{$\tan(\theta)+\sec(\theta)+C$}$
