integration - Evaluate $int frac{1+cos(x)}{sin^2(x)},operatorname d!x$

I`m trying to solve this integral and I did the following steps to solve it but don't know how to continue.
$$\int \dfrac{1+\cos(x)}{\sin^2(x)}\,\operatorname d\!x$$

$$\begin{align}\int \dfrac{\operatorname d\!x}{\sin^2(x)}+\int \frac{\cos(x)}{\sin^2(x)}\,\operatorname d\!x &= \int \dfrac{\operatorname d\!x}{\sin^2(x)}+\int \frac{\cos(x)}{1-\cos^2(x)} \\ &=\int \sin^{-2}(x)\,\operatorname d\!x + \int \cos(x)\,\operatorname d\!x - \int \frac{\operatorname d\!x}{\cos(x)}\end{align}$$
Any suggestions how to continue?

Thanks!