I like this integral

$\begin{aligned} I & = \int_0^\frac \pi 4 \frac {\cos x}{\sin x+\cos x}dx & \small \color{#3D99F6} \text{Using }\int_a^b f(x) \ dx = \int_a^b f(a+b-x) \ dx \\ & = \int_0^\frac \pi 4 \frac {\cos \left(\frac \pi 4-x\right)}{\sin \left(\frac \pi 4-x\right)+\cos \left(\frac \pi 4-x\right)} dx \\ & = \int_0^\frac \pi 4 \frac {\frac 1{\sqrt 2} \cos x + \frac 1{\sqrt 2}\sin x}{\frac 1{\sqrt 2} \cos x - \frac 1{\sqrt 2}\sin x + \frac 1{\sqrt 2} \cos x + \frac 1{\sqrt 2}\sin x} dx \\ & = \int_0^\frac \pi 4 \frac {\cos x + \sin x}{2\cos x} dx \\ & = \frac 12 \int_0^\frac \pi 4 \left(1+\tan x \right) dx \\ & = \frac {x - \ln (\cos x)}2 \bigg|_0^\frac \pi 4 \\ & = \frac \pi 8 + \frac {\ln 2}4 \\ & \approx \boxed{0.566} \end{aligned}$

The fact that the derivative of $\sin(x) + \cos(x)$ is $\cos(x) - \sin(x)$ is the motivation behind this method. So note that

$\dfrac{\cos(x)}{\sin(x) + \cos(x)} = \dfrac{1}{2}\left(\dfrac{2\cos(x)}{\sin(x) + \cos(x)}\right) = \dfrac{1}{2}\left(\dfrac{\cos(x) + \cos(x) + \sin(x) - \sin(x)}{\sin(x) + \cos(x)}\right) =$

$\dfrac{1}{2}\left(\dfrac{\sin(x) + \cos(x)}{\sin(x) + \cos(x)} + \dfrac{\cos(x) - \sin(x)}{\sin(x) + \cos(x)}\right) = \dfrac{1}{2}\left(1 + \dfrac{\cos(x) - \sin(x)}{\sin(x) + \cos(x)}\right)$ .

Then using the substitution method with $u = \sin(x) + \cos(x) \Longrightarrow du = (\cos(x) - \sin(x)) dx$ , we see that

$\displaystyle \int_{0}^{\pi/4} \dfrac{\cos(x)}{\sin(x) + \cos(x)} dx = \dfrac{1}{2}\left(x + \ln(\sin(x) + \cos(x))\right)_{0}^{\pi/4} = \dfrac{1}{2}\left(\dfrac{\pi}{4} + \ln\left(\dfrac{1}{\sqrt{2}} + \dfrac{1}{\sqrt{2}}\right)\right) - \dfrac{1}{2}(0 + \ln(0)) = \dfrac{\pi}{8} + \dfrac{\ln(2)}{4} = \boxed{0.566}$ to 3 decimal places.

$\begin{aligned} \int_0^{\frac{\pi}{4}} \frac{\cos(x)}{\sin(x)+\cos(x)}\,dx= & \int_0^{\frac{\pi}{4}} \frac{\cos (x) (\sin(x) + \cos(x))}{(\sin(x)+\cos(x))^2}\,dx\\ =& \int_0^{\frac{\pi}{4}} \frac{\cos(x)\sin(x)+\cos^2(x)}{1+2\cos(x)\sin(x)}\,dx\\ = &\frac{1}{2} \int_0^{\frac{\pi}{4}}\frac{\sin(2x)+\cos(2x)+1}{1+\sin(2x)}\,dx \\ = & \frac{1}{2}\int_0^{\frac{\pi}{4}}1 + \frac{\cos(2x)}{1+\sin(2x)}\,dx \\ = & \frac{\pi}{8} +\frac{1}{2}\int_0^{\frac{\pi}{4}} \frac{\cos(2x)}{1+\sin(2x)}\,dx \\ = & \frac{\pi}{8} + \frac{1}{4}\int_0^1 \frac{1}{1+u}\, du \, \, \text{ via } \color{#3D99F6} u = \sin(2x)\\ = & \frac{\pi}{8} + \frac{1}{4} \ln(2) \approx \boxed{0.565985} \end{aligned}$

Write $I_1 = \int_0^{\frac{\pi}{4}} \frac{\cos(x)}{\sin(x)+\cos(x)}\mathrm{d}x \text{ and } I_2 = \int_0^{\frac{\pi}{4}} \frac{\sin(x)}{\sin(x)+\cos(x)}\mathrm{d}x.$ Note that $I_1 + I_2 = \frac{\pi}{4}$ . Furthermore, we can observe that $I_1 - I_2 = \int_0^{\frac{\pi}{4}} \frac{\cos(x)-\sin(x)}{\sin(x)+\cos(x)}\mathrm{d}x.$ Substituting $u = \sin(x) + \cos(x)$ implies $\mathrm{d}u = \left(\cos(x)-\sin(x)\right)\mathrm{d}x$ from which follows that $I_1 - I_2 = \int_1^{\sqrt{2}} \frac{\mathrm{d}u}{u} = \frac{\ln(2)}{2}.$ By using the two equations $I_1 + I_2 = \frac{\pi}{4}$ and $I_1 - I_2 = \frac{\ln(2)}{2}$ we obtain that $I_1 = \frac{\pi}{8} + \frac{\ln(2)}{4} \approx \boxed{0.566}.$

We can use the identity $\cos(x-\phi) = \cos(x)\cos(\phi) + \sin(x)\sin(\phi)$ and choose $\phi = \frac{\pi}{4}$ to rewrite the denominator as $\sin x + \cos x = \sqrt{2} \cos\left(x - \frac{\pi}{4}\right).$

Then use the substitution $u = x-\frac{\pi}{4}$ to find the equivalent integral $\begin{aligned} \int_{-\pi/4}^0 \frac{\cos\left(u+\frac{\pi}{4}\right)}{\sqrt{2}\cos u} \,du &= \int_{-\pi/4}^0 \frac{\frac{\sqrt{2}}{2}\cos u - \frac{\sqrt{2}}{2}\sin u}{\sqrt{2}\cos u}\, du \\ &= \frac{1}{2} \int_{-\pi/4}^0 \left(1 - \tan u\right)\,du \\ &= \frac{1}{2} \Big[u + \ln|\cos u|\Big]_{-\pi/4}^0 \\ &= \frac{1}{2} \left( (0 + \ln(1)) - \left(-\frac{\pi}{4} + \ln\left(\frac{1}{\sqrt{2}} \right) \right) \right) \\ &= \frac{\pi}{8} + \frac{1}{4}\ln(2) \approx \boxed{0.565985877} \end{aligned}$