Answer
$\ln|1+\sin\theta|+C$
Work Step by Step
$\int\frac{1}{\sec\theta+\tan\theta}d\theta=\int\frac{1}{\frac{1}{\cos\theta}+\frac{\sin\theta}{\cos\theta}}d\theta=\int \frac{\cos\theta}{1+\sin\theta}d\theta$
With $u=\sin\theta$ so that $d\theta=\frac{du}{\cos\theta}$, we have
$\int\frac{\cos\theta}{1+\sin\theta}d\theta=\int\frac{\cos\theta}{1+u}\frac{du}{\cos\theta}=\int\frac{1}{1+u}du$
With v=1+u so that du=dv, we have
$\int\frac{1}{1+u}du=\int\frac{1}{v}dv=\ln|v|+C=\ln|1+u|+C=\ln|1+\sin\theta|+C$
