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
 A: 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 $\sin(x)$ [$\cos^2x+4=5-\sin^2 x$]
A: 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 $$\frac{\sqrt{5}}{2}\arctan\left(\frac{2}{\sqrt{5}}\tan x\right)-x+c$$
A: 
$$\int \frac{1}{\cos^2x +4}dx$$

Divide top and bottom of integrand by $\cos^2 x$,
$$=\int \frac{\sec^2 x}{1+4\sec^2 x} dx $$
Now use $\tan^2 x+1=\sec^2 x$ to get,
$$=\int \frac{\sec^2 x}{1+4(\tan^2 x+1)} dx$$
$$=\int \frac{\sec^2 x}{5+4 \tan^2 x} dx$$
Let $u=\tan x$.
$$=\int \frac{1}{5+4u^2} du$$
$$=\frac{1}{5} \int \frac{1}{1+\frac{4}{5}u^2} du$$
$$=\frac{1}{5} \int \frac{1}{1+(\frac{2}{\sqrt{5}}u)^2} du$$
$$=\frac{1}{5} \frac{\sqrt{5}}{2} \int \frac{\frac{2}{\sqrt{5}}}{1+(\frac{2}{\sqrt{5}}u)^2} du$$
I think you can deal with this.
