Given the following differential equation $$y''(t)+4y(t)=\frac{1}{4+\cos{2t}}$$ with $y(0)=1$, $y'(0)=0$, find $y(t)$.
Using the Laplace Transform, I've got that $$Y(s)=\frac{s}{s^2+4}+\frac{F(s)}{s^2+4}$$ where $F$ is the Laplace transform of $\frac{1}{4+\cos{2t}}$. One can immediately observe how the first term is the image of $\cos{2t}$. But what about the second one? If I use the inverse Laplace Transform of the product $\cfrac{F(s)}{s^2+4}$, I have to compute the convolution between $\cos{2t}$ and $\cfrac{1}{4+\cos{2t}}$, which is $$\int_0^t \frac{\sin(2t-2u)}{4+\cos(2u)}\,du$$
Now, I could use the fact that $\sin(a-b)=\sin a\cos b-\sin b\cos a$.
I was wondering, do I really have to compute this integral? I could also use the variation of parameters method, but I would run into the same integral.