Multiply the integrated function by $\displaystyle \frac{1+sin(x)}{1+sin(x)}$
With some algebra and trigonometric identities, you will get:
$\displaystyle \int \frac{sin^2(x)+2sin(x)+1}{cos^2(x)} dx$
$\displaystyle =\int tan^2(x)dx + 2 \int \frac{sin(x)}{cos^2(x)} dx + \int sec^2(x) dx$
For the first: use a well-known trigometric identity.
For the second: use a U-Substitution.
For the third: Its a well-known integral.
$\displaystyle \int\frac{1+\sin(x)}{1-\sin(x)}\text{ }dx=\int\frac{1+\sin(x)}{1-\sin(x)}\cdot\frac{1+\sin(x)}{1+\sin(x)}\text{ }dx=$$\displaystyle \int\frac{\left(1+\sin(x)\right)^2}{1-\sin^2(x)}\text{ }dx=\int\frac{\left(1+\sin(x)\right)^2}{\cos^2(x)} \text{ }dx$. I'm sure you can go from there.