Solving differential equation using a pair of integrals as helper I found an interesting problem, but I could only solve half of it:
Given $$ I_{1} = \int \frac{e^{-x}  + \sin{x}}{e^{-x}  + \sin{x}  + \cos{x}} dx \\  I_{2} = \int \frac{\cos{x}}{e^{-x}  + \sin{x}  + \cos{x}} dx  $$ 
and $f:(0,\frac{\pi}{2})\rightarrow \Bbb R$ and one of it's primitives $F:(0,\frac{\pi}{2})\rightarrow \Bbb R $, find $f(x)$ from this equation:
$(e^{-x}  + \sin{x}  + \cos{x})F(x)=\cos{}x-x(e^{-x}  + \sin{x}  + e^{x}\cos{x})f(x), x \in (0,\frac{\pi}{2})$.
I managed to solve the integrals: $$I_{1} = \frac{1}{2}x+\frac{1}{2}\ln{|e^{-x}  + \sin{x}  + \cos{x})|} \\I_{2} = \frac{1}{2}x-\frac{1}{2}\ln{|e^{-x}  + \sin{x}  + \cos{x})|} $$ 
But I cannot figure out the link between these two answers and the functional equation. I tried solving it as a first order linear differential equation (knowing that $\frac{d}{dx}F(x)=f(x)$), but the final answer is very complicated. Can you help me figure out what the answer to the equation is? Thanks in advance!

for your second integral: substitute $$t=e^x\sin(x)+e^x\cos(x)+1$$ then we get
$$dt=2e^x\cos(x)dx$$ and the integral will be $$\frac{1}{2}\int \frac{1}{t}dt$$
and at first you can write $$\frac{\cos(x)}{e^{-x}+\sin(x)+\cos(x)}=\frac{e^x\cos(x)}{1+e^x\sin(x)+e^x\cos(x)}$$
