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

HINT:
$$\frac{x^2}{5-x^2}=-\frac{-x^2+5-5}{5-x^2}=-1+\frac{5}{5-x^2}=-1+\frac{A}{\sqrt{5}-x}+\frac{B}{\sqrt{5}+x}\\A(\sqrt{5}+x)+B(\sqrt{5}-x)=5$$
Now when you solve for $A,B$ instead of $x$ plug in $\sin(x)$ [$\cos^2x+4=5-\sin^2 x$]
