Solve this nonhomegenous ode $y''+4y = \cos(2x)$ Solve the given nonhomogenous linear ODE by variation of parameters or undetermined coefficients $y'' + 4y = \cos(2x)$.
A general solution is $y_1 = \cos(2x), y_2 = \sin(2x)$, and the Wronksin determinant is equal to 2.
Plugging this into the equation for the method of variation of parameters, I get
$$-\cos(2x)\int\frac{\sin(2x)\cos(2x)}{2}dx + \sin(2x)\int\frac{\cos^2(2x)}{2}dx$$
The integrals cancel out to $0$. How can I approach this correctly with the method proposed?

Null space solution
$m^2 +4 =0$
$\implies m=\pm 2i$
Hence,  $ y_N = C_1 \sin 2x + C_2 \cos 2x$.
Since, $\cos2x$ already in null space solution, take particular integral as $$y_p =x(A \sin 2x +B \cos 2x) $$
Then, \begin{align}  y'_p &{=A \sin 2x +B \cos 2x +2x(A \cos 2x -B \sin 2x )}\\ ​\end{align}
$y''_p =4A \cos 2x -4B \sin 2x -4x(A\sin 2x + B \cos 2x )$
$$y''_p +4y'_p = \cos 2x$$
\begin{align} {4A\cos 2x -4B\sin 2x -4x(A\sin 2x + B\cos 2x )+4x(A \sin 2x +B \cos 2x) =\cos 2x } \end{align}
$4A\cos 2x -4B\sin 2x= \cos 2x$
$4A=1$ , $4B=0$
$A=1/4$ , $B=0$
Hence, $y_p = \frac {1}{4}{x\sin 2x }$
So, $$y_\text{complete} =y_N + y_p$$
$$y_\text{complete}=C_1\sin 2x + C_2\cos 2x +  \frac {1}{4}{x \sin 2x }$$
