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