Primitive of an Exact form I have the following differential $3$-form,
$$ H = \cos \zeta \sin \zeta\, \mathrm{d}\zeta \wedge \mathrm{d}\varphi_1 \wedge \mathrm{d}\varphi_2 $$
which I know is exact by the properties of my problem.
Now I would like to find a primitive form $B$ such that $H = \mathrm{d}B$.
I figured out it will depend on the cosine of the difference $\varphi_1 -\varphi_2$ (such that the derivative with respect the two angles will disappear at the end. But I don't manage to do the formal complete calculation. Any idea?
 A: You may set:
$$B = f(\zeta ,{\varphi _1},{\varphi _2}){\kern 1pt} {\text{d}}{\varphi _1} \wedge {\text{d}}{\varphi _2}$$
Then exterior derivative gives:
$$\begin{gathered}
  dB = df(\zeta ,{\varphi _1},{\varphi _2}){\kern 1pt}  \wedge {\text{d}}{\varphi _1} \wedge {\text{d}}{\varphi _2} \hfill \\
   = (\frac{{\partial f}}{{\partial \zeta }}{\text{d}}\zeta  + \frac{{\partial f}}{{\partial {\varphi _1}}}{\text{d}}{\varphi _1} + \frac{{\partial f}}{{\partial {\varphi _2}}}{\text{d}}{\varphi _2}) \wedge {\text{d}}{\varphi _1} \wedge {\text{d}}{\varphi _2} \hfill \\
   = \frac{{\partial f}}{{\partial \zeta }}{\text{d}}\zeta  \wedge {\text{d}}{\varphi _1} \wedge {\text{d}}{\varphi _2} \hfill \\ 
\end{gathered} $$
That means:
$$\frac{{\partial f}}{{\partial \zeta }}(\zeta ,{\varphi _1},{\varphi _2}) = \cos \zeta \sin \zeta $$
So:
$$f(\zeta ,{\varphi _1},{\varphi _2}) =  - \frac{1}{2}{\cos ^2}(\zeta )$$
and
$$B =  - \frac{1}{2}{\cos ^2}(\zeta ){\text{d}}{\varphi _1} \wedge {\text{d}}{\varphi _2}$$
with:
$$H = dB = \cos \zeta \sin \zeta {\kern 1pt} {\text{d}}\zeta  \wedge {\text{d}}{\varphi _1} \wedge {\text{d}}{\varphi _2}$$
More general:
If we set:
$$A = f \cdot {\text{d}}{\varphi _1} \wedge {\text{d}}{\varphi _2} + g \cdot {\kern 1pt} {\text{d}}\zeta  \wedge {\text{d}}{\varphi _2} + h{\kern 1pt}  \cdot {\text{d}}\zeta  \wedge {\text{d}}{\varphi _1}$$
then:
$$\begin{gathered}
  dA = \frac{{\partial f}}{{\partial \zeta }}{\text{d}}\zeta  \wedge {\text{d}}{\varphi _1} \wedge {\text{d}}{\varphi _2} + \frac{{\partial g}}{{\partial {\varphi _1}}}{\text{d}}{\varphi _1} \wedge {\text{d}}\zeta  \wedge {\text{d}}{\varphi _2} + \frac{{\partial h}}{{\partial {\varphi _2}}}{\text{d}}{\varphi _2} \wedge {\text{d}}\zeta  \wedge {\text{d}}{\varphi _1} \hfill \\
  dA = (\frac{{\partial f}}{{\partial \zeta }} + \frac{{\partial h}}{{\partial {\varphi _2}}} - \frac{{\partial g}}{{\partial {\varphi _1}}}){\text{d}}\zeta  \wedge {\text{d}}{\varphi _1} \wedge {\text{d}}{\varphi _2} \hfill \\ 
\end{gathered}$$
Comparing gives:
$$\frac{{\partial f}}{{\partial \zeta }} + \frac{{\partial h}}{{\partial {\varphi _2}}} - \frac{{\partial g}}{{\partial {\varphi _1}}} = \cos \zeta \sin \zeta $$
So we conclude:
$$\begin{gathered}
  \frac{{\partial f}}{{\partial \zeta }} = \cos \zeta \sin \zeta  \hfill \\
  \frac{{\partial h}}{{\partial {\varphi _2}}} = \frac{{\partial g}}{{\partial {\varphi _1}}} \hfill \\ 
\end{gathered}$$
Now setting:
$$\begin{gathered}
  h = \frac{{\partial P}}{{\partial {\varphi _1}}} \hfill \\
  g = \frac{{\partial P}}{{\partial {\varphi _2}}} \hfill \\
  f(\zeta ,{\varphi _1},{\varphi _2}) =  - \frac{1}{2}{\cos ^2}(\zeta ) \hfill \\ 
\end{gathered}$$
for a function $P$ we have:
$$\begin{gathered}
  A =  - \frac{1}{2}{\cos ^2}(\zeta ){\text{d}}{\varphi _1} \wedge {\text{d}}{\varphi _2} - \left( {\frac{{\partial P}}{{\partial {\varphi _1}}}{\text{d}}{\varphi _1} + \frac{{\partial P}}{{\partial {\varphi _2}}}{\text{d}}{\varphi _2}} \right) \wedge {\text{d}}\zeta  \hfill \\
  A =  - \frac{1}{2}{\cos ^2}(\zeta ){\text{d}}{\varphi _1} \wedge {\text{d}}{\varphi _2} - \left( {\frac{{\partial P}}{{\partial \zeta }}{\text{d}}\zeta  + \frac{{\partial P}}{{\partial {\varphi _1}}}{\text{d}}{\varphi _1} + \frac{{\partial P}}{{\partial {\varphi _2}}}{\text{d}}{\varphi _2}} \right) \wedge {\text{d}}\zeta  \hfill \\
  A =  - \frac{1}{2}{\cos ^2}(\zeta ){\text{d}}{\varphi _1} \wedge {\text{d}}{\varphi _2} - {\text{d}}P \wedge {\text{d}}\zeta  \hfill \\
  A =  - \frac{1}{2}{\cos ^2}(\zeta ){\text{d}}{\varphi _1} \wedge {\text{d}}{\varphi _2} + {\text{d}}\zeta  \wedge {\text{d}}P \hfill \\ 
\end{gathered} $$
That is:
$$B =  - \frac{1}{2}{\cos ^2}(\zeta ){\text{d}}{\varphi _1} \wedge {\text{d}}{\varphi _2} = A - {\text{d}}\zeta  \wedge {\text{d}}P$$
$B$ differs from $A$ by a closed form ${\text{d}}\zeta  \wedge {\text{d}}P$
A: Why not trying to write something like
\begin{equation}
B=B_{1} (\zeta,\varphi_1,\varphi_2)\mathrm{d}\varphi_1 \wedge \mathrm{d}\varphi_2+B_{2}(\zeta,\varphi_1,\varphi_2)\mathrm{d}\zeta \wedge \mathrm{d}\varphi_1+B_{3}(\zeta,\varphi_1,\varphi_2)\mathrm{d}\zeta \wedge \mathrm{d}\varphi_2
\end{equation}
If $H=dB$ is exact, you may find constraints on each $B_{i}$.
A: In general, assume $f$ is a continuous function of one variable and $F$ is an antiderivative of $f$ (i.e., $F' = f$). If 
$$
H = f(x)\, dx \wedge dy,\qquad
B = F(x)\, dy,
$$
then $dB = H$.
Similar observations hold for forms of higher degree, and the $3$-form
$$
H = \cos\zeta \sin\zeta\, d\zeta \wedge d\varphi_{1} \wedge d\varphi_{2}
$$
is of this type. (Particularly, $H$ is exact a priori, by inspection. :)
