Failure to correctly apply the method of Variation of Parameters I use the Method of Variation of Parameters to find a particular solution to a given linear inhomogeneous ODE. Subsequently, I insert that particular solution and its derivatives into the inhomogeneous ODE. I find that, contradictorily, the particular solution is infact no solution. If there is somebody to tell me where there is an error in either my arithmetic or my logic, then I will thank you very much.
$$  y'' + y = \frac{1}{\cos x +1}  $$
$$ y = y_p + y_h $$
$$ y_h = \, c_1 \cos x + c_2 \sin x  $$
$$ y_p = \int_{x_0}^x \frac{\mathrm W_1 (t)}{\mathrm{W} (t)} \, \mathrm d t \, \cos x + \int_{x_0}^x \frac{\mathrm W_2 (t)}{\mathrm{W} (t)} \, \mathrm d t \, \sin x $$
$$ \mathrm W (t) = \left | 
  \begin{matrix}
  \cos t & \sin t  \\
  - \sin t & \cos t  
  \end{matrix}
  \right |  
  = 1 $$
$$
   \mathrm W_1 (t) = \left | 
  \begin{matrix}
  0 & \sin t  \\
  (1 + \cos t)^{-1} & \cos t  
  \end{matrix}
  \right |  
  = \frac{-\sin t}{1+\cos t} \\
$$
$$
  \mathrm W_2 (t) = \left | 
  \begin{matrix}
  \cos t & 0  \\
  - \sin t  & (1 + \cos t)^{-1}  
  \end{matrix}
  \right |  
  = \frac{\cos t}{1+\cos t}
$$
$$
\int_{x_0}^x \frac{-\sin t}{1+\cos t} \, \mathrm d t \, = \ln {\left | 1 + \cos x \right |}
$$
$$
\int_{x_0}^x \frac{\cos t}{1+\cos t} \, \mathrm d t \, = x - \tan \frac {x}{2}
$$
$$
\begin{align*}
  y_p =& \cos x \ln |1 + \cos x| + \sin x \left( x - \tan \frac x 2 \right) \\
      =& \cos x \ln |1 + \cos x| + x \sin x  - (1 - \cos x)
\end{align*}
$$
$$
\begin{align*}
y'_p =& - \sin x \ln \left | 1 + \cos x \right | - \cos x (1 + \cos x)^{-1} \sin x + \sin x + x \cos x - \sin x \\
     =& - \sin x \ln \left | 1 + \cos x \right | - \cos x (1 - \cos x) \sin^{-1} x + x \cos x  \\
     =& - \sin x \ln \left | 1 + \cos x \right | - \cot x + \cos^2 x \sin^{-1} x + x \cos x  \\     
\end{align*}
$$
$$
\begin{align*}
y''_p =& -\cos x \ln |1 + \cos x| + \sin^2 x (1+ \cos x)^{-1} + \sin^{-2} x - 2 \cot x \sin x \\
       & \,\,\,\, -\cot^2 x \cos x + \cos x - x \sin x
\end{align*}
$$
Note that $  \sin^2 x (1+ \cos x)^{-1} = 1-\cos x  $
$$ y''_p = -\cos x \ln |1 + \cos x | + 1 + \sin^{-2} x - 2 \cos x   -\cot^2 x \cos x - x \sin x $$
$$
\begin{align*}
y''_p + y_p =&  \sin^{-2} x -\cos x -\cot^2 x \cos x \\
=& - \cos x + \sin^{-2} x \, (1- \cos^3 x) \\
=& - \cos x + \sin^{-2} x \, (1- \frac{1}{4} \cos 3x + \frac{3}{4} \cos x) \\
=& - \cos x + \sin^{-2} x \, (1- \frac{1}{4} (\cos 2x \cos x - \sin 2x \sin x) + \frac{3}{4} \cos x) \\
=& - \cos x + \sin^{-2} x - \frac{1}{4} \cot x \sin^{-1} x \cos 2x + \frac{1}{4} \sin^{-1} x \sin 2x + \frac{3}{4} \cot x \sin^{-1} x \\
=& - \cos x + \sin^{-2} x + \frac{1}{4} \sin^{-1} x \sin 2x + \frac{1}{4} \sin^{-1} x \cot x (3 - \cos 2x) \\
\end{align*}
$$
Note that $ 3 - \cos 2x = 2+1 - \cos 2x = 2 + 2 \sin^2 x = 2 (1 + \sin^2 x) $
And that $ \sin 2x = 2 \sin x \cos x \Longrightarrow \frac{1}{4} \sin^{-1} x \sin 2x = \frac{1}{2} \cos x $
$$
\begin{align*}
y''_p + y_p =& - \cos x + \sin^{-2} x + \frac{1}{2} \cos x + \frac{1}{2} \cot x \sin^{-1} x  (1 + \sin^2 x) \\
=& - \frac{1}{2} \cos x + \sin^{-2} x + \frac{1}{2} \cos x \sin^{-2} x + \frac{1}{2} \cos x  \\
=& \sin^{-2} x (1+\frac{1}{2} \cos x) \\
\neq& \sin^{-2} x (1- \cos x) \\
=& \frac{1}{\cos x +1} 
\end{align*}
$$
If the arithmetic that I have shown here is correct, then the function $y_p$ that I obtained by Variation of Parameters is not a solution to the given inhomogeneous ODE.

Here is another method that help check your answer
Without Wronskian method
$$y'' + y = \frac{1}{\cos x +1}$$
$$\sin(x) y'' + y\sin(x) = \frac{\sin(x)}{\cos x +1}$$
$$\sin(x) y'' +y'\cos(x)-y'\cos(x)+ y\sin(x) = \frac{\sin(x)}{\cos x +1}$$
$$(\sin(x) y')'+(-y\cos(x))' = \frac{\sin(x)}{\cos x +1}$$
Integrate
$$\sin(x) y'+(-y\cos(x)) =  \int \frac{\sin(x)}{\cos x +1}dx$$
$$\sin(x) y'-y\cos(x)=  -\ln|\cos x +1|+K_1$$
Which is of first order
$$\frac y {\sin(x)}=-K_1 \cot(x)  -\int \frac {\ln|\cos x +1|}{\sin^2(x)}dx$$
$$\frac y {\sin(x)}= -K_1 \cot(x) + \cot(x){\ln|\cos x +1|} +\int \frac {\cos(x)}{\cos x +1|}dx$$
$$ y = C_1 \cos(x)  +\cos(x){\ln|\cos x +1|} +\sin(x)\int \frac {\cos(x)}{\cos x +1|}dx+K_2\sin(x)$$
$$ \boxed{y = C_1 \cos(x) + K_2\sin(x)+\cos(x){\ln|\cos x +1|} +\sin(x) ( x-\tan(x/2))}$$

Edit
You made a sign mistake when you linearised $\cos^3(x)$
$$cos^3(x)=\cos(3x)/4+3\cos(x)/4$$
You wrote this
$$\begin{align*}
y''_p + y_p =&  \sin^{-2} x -\cos x -\cot^2 x \cos x \\
=& - \cos x + \sin^{-2} x \, (1- \cos^3 x) \\
=& - \cos x + \sin^{-2} x \, (1- \frac{1}{4} \cos 3x \color{red}{+ \frac{3}{4} \cos x)} \\
\end{align*}
$$
It should be
$$y''_p + y_p = - \cos x + \sin^{-2} x \, (1- \frac{1}{4} \cos 3x \color{blue}{- \frac{3}{4} \cos x)} $$
