Need direction for solving integral 
Evaluate: $$\int \frac{\sin^2x}{\cos^2x +4}dx.$$

I tried to this things:
First
$$\tan\left(\frac{x}{2}\right) = t$$
$$dx = \frac{2~dt}{1+t^2}$$
$$\sin x= \frac{2t}{1+t^2}$$
$$\cos x= \frac{1-t^2}{1+t^2}$$
Tried also but this is same thing?
$$\tan\left(\frac{x}{2}\right) = t$$
There I don't have to place $2$ left to integral ( same thing )
Second
I tried trigonometric transformations like
$$\sin^2x \equiv 1-\cos^2x$$
$$\int \frac{1-\cos^2x}{\cos^2x +4}dx$$
$$-1\cdot \int \frac{\cos^2x-1}{\cos^2x +4}~dx$$
$$-1\cdot \int \frac{\cos^2x+4-5}{\cos^2x +4}~dx$$
But I get stucked in this part
$$5\int \frac{1}{\cos^2x +4}dx.$$
And mix something from first step
Third
I tried to divide everything with $\sin^2x$, but didn't succeed to solve + mix something from first step
Can anyone give me direction how to solve this?
sorry, for late update....
My goal is to complete this task without trigonometry function sec

substituting $t=\tan x$, with $dx=\frac{dt}{1+t^2}$ gives $$\int\frac{t^2}{(5+4t^2)(1+t^2)}dt=\int\frac{5}{5+4t^2}-\frac{1}{1+t^2}dt$$
So you end up with $$\frac{\sqrt{5}}{2}\arctan\left(\frac{2}{\sqrt{5}}\tan x\right)-x+c$$
