${\int }_0^{\frac{\pi }{4}} \frac{{\cos }^{2}x}{{\cos }^{2}x+4{\sin }^{2}x}dx=$

$={\int }_0^{\frac{\pi }{4}} \frac{{\cos }^{2}x}{{\cos }^{2}x+4{\sin }^{2}x}dx$

$={\int }_0^{\frac{\pi }{4}} \frac{1}{1+4{\tan }^{2}x}dx$

$={\int }_0^{\frac{\pi }{4}} \frac{se{c}^{2}x}{se{c}^{2}x\left(1+4{\tan }^{2}x\right)}dx$

Put

$$\begin{gathered} \tan x=t \\ se{c}^{2}xdx=dt \end{gathered}$$

$={\int }_0^1 \frac{dt}{\left(1+t^2\right)\left(1+4t^2\right)}$

$={\int }_0^1 \left[\frac{A}{1+t^2}+\frac{B}{1+4t^2}\right]dt$

$\begin{array}{r}1=A\left(1+4t^2\right)+B\left(1+t^2\right)\\ A+B=1 ,4A+B=0\end{array}$

$\begin{array}{r}A-4A=1\hspace{0.25em},\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}B=-4A\\ A\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=-1/3,\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}B\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=4/3\end{array}$

$I=-\frac{1}{3}{\int }_0^1 \frac{dt}{1+t^2}+2/3{\int }_0^1 \frac{2}{1+(2t{)}^2}dt$

$=-\frac{\pi }{12}+\frac{2}{3}{\mathrm{Tan}}^{-1}2$