So I used variation of parameter and concluded that the particular solution follows:

$$y_p = 2\cos (2t) \int{\tan (t) \sin (2t) dt} + 2 \sin (2t) \int{\tan(t) \cos(2t) dt}$$

I am not sure how to solve the two integrals $\int{\tan (t) \sin (2t) dt}$ and $\int{\tan(t) \cos(2t) dt}$ since it involves the $2t$ term inside the sine and cosine function. Can anyone help me solve this?

First, recall the double-angle trig identity, $\sin(2t)=2\sin(t)\cos(t)$. Thus, the integrand $\tan(t)\sin(2t)=\tan(t)(2\sin(t)\cos(t))=2\sin^2(t)$.

The half-angle trig identity allows us to rewrite $2\sin^2(t)$ as $1-\cos(2t)$.

Now, we have

$$\int{1}dt-\int{\cos(2t)}dt$$

Let $u=2t \Rightarrow du=2dt$

$$\int{1}dt-\int{\cos(2t)}dt=t-\frac{1}{2}\int{\cos(u)}du=t-\frac{1}{2}\sin(u)=t-\frac{1}{2}\sin(2t)=t-\sin(t)\cos(t)$$

The process is similar for $\int{\tan(t)\cos(2t)}dt$. Hope that helps!