Finding a generating function for a canonical transformation

[fluidistic]

Homework Statement

I'm trying to find a generating function for the canonical transformation Q=\left ( \frac{\sin p}{q} \right ), P=q \cot p.

Homework Equations

I am not really sure. I know there are 4 different types of generating function. I guess it's totally up to me to choose the type of the generating function I want, since it's not specified in the problem statement?
So let's say I want a type 1. Thus F_1 depends on q and Q.
\frac{\partial F_1}{\partial q}=p
\frac{\partial F_1}{\partial Q}=-P.

The Attempt at a Solution

From the relevant equations, \partial F_1=-P \partial Q and \partial F_1=p \partial q.
I don't really know why, but apparently dF_1=pdq-PdQ. It looks like dF_1=2 \partial F_1 for some reason unknown to me.
Now I think the main idea to get F_1 is to express dF_1 as a single differential and then integrate.
I have that p= \cot ^{-1} \left ( \frac{P}{q} \right ) and dQ=\frac{1}{\sin p } \left [ \frac{\cos p}{q}dp-\frac{\sin p}{q^2}dq \right ].
This gives me dF_1=(p+ \cot p)dq-q\cot ^2 p dp. In a book I found that this is equal to d[q(p+\cot p)] (how?!), therefore F_1=q \cot ^{-1} \left ( \frac{P}{q} \right )+P. But it seems that F depends on P and q instead of on q and Q. So I "found" a type 2 generating function instead of type 1... I think I know why, I found p instead of q. Anyway my problems remains.
1)Why is dF=pdq-PdQ?
2)Why and how do you figure out that (p+ \cot p)dq-q\cot ^2 p dp=d[q(p+\cot p)]?
I'd be very grateful if someone could answer these 2 questions.
Thanks!

Edit: I just answered the "why" of my question 2. I just don't know how do one figures this out.

Edit 2: I do not know why I found out a type 2, namely F(P,q) while I was looking for F(q,Q).

 

[member 553083]

I guess I "transformed" the problem back to the original one, before the transformation. Now I understand why dF=pdq-PdQ, that's because of the transformation equations. I guess now I just have to figure out the answer to my question 1. How do you figure out that (p+ \cot p)dq-q\cot ^2 p dp=d[q(p+\cot p)]?