Answer
The solution is $$\int\frac{4}{\csc\theta-\cot\theta}d\theta=4\ln|1-\cos\theta|+c.$$
Work Step by Step
The integral $$\int\frac{4}{\csc\theta-\cot\theta}d\theta$$ can be transformed into:
$$\int\frac{4}{\csc\theta-\cot\theta}d\theta=\int\frac{4}{\frac{1}{\sin\theta}-\frac{\cos\theta}{\sin\theta}}d\theta=\int\frac{4}{\frac{1-\cos\theta}{\sin\theta}}d\theta=\int\frac{4\sin\theta}{1-\cos\theta}d\theta.$$
Now we can use substitution $1-\cos\theta=t\Rightarrow\sin\theta d\theta=dt$ and put that into our integral:
$$\int\frac{4\sin\theta}{1-\cos\theta}d\theta=4\int\frac{dt}{t}=4\ln|t|+c,$$
where $c$ is arbitrary constant. Now we have to express our solution (which is in terms of $t$) in terms of $\theta$:
$$\int\frac{4\sin\theta}{1-\cos\theta}d\theta=4\ln|t|+c=4\ln|1-\cos\theta|+c.$$
