The easiest way to solve the following integral : $\int_{\pi/6}^{\pi/3} \frac{\sin x}{\sin x+\cos x} \mathrm{d}x$ Today I passed the test to access to medicine, and I found some really cool questions on the mathematics part, anyways I just went to see the difficulty of the test, and sorry for being far from my main question :
$$\int_{\pi/6}^{\pi/3} \frac{\sin x}{\sin x+\cos x} \mathrm{d}x$$
I used (obviously) the king property of integration which gives $\pi/12$ as an answer.
My question is how can we solve this using some basic properties of integration ?
Thanks in advance. and please wish me some luck for the next tests to access engineering universities.

HINT
Make the change of variable $y = \pi/2 - x$. Then we get that
\begin{align*}
I & =
\int_{\pi/6}^{\pi/3}\frac{\sin(x)}{\sin(x) + \cos(x)}\mathrm{d}x = \int_{\pi/6}^{\pi/3}\frac{\cos(y)}{\cos(y) + \sin(y)}\mathrm{d}y
\end{align*}
Then add both expressions together and see what happens.
Can you take it from there?
EDIT
To begin with, notice that
\begin{align*}
\frac{\sin(x)}{\sin(x) + \cos(x)} & = \frac{1}{2}\left(\frac{2\sin(x)}{\sin(x) + \cos(x)}\right)\\\\
& = \frac{1}{2}\left[\frac{(\sin(x) + \cos(x)) + (\sin(x) - \cos(x))}{\sin(x) + \cos(x)}\right]\\\\
& = \frac{1}{2} - \frac{1}{2}\left(\frac{\cos(x) - \sin(x)}{\sin(x) + \cos(x)}\right)
\end{align*}
Moreover, we do also have that $(\sin(x) + \cos(x))' = \cos(x) - \sin(x)$.
Having said that, according to the substitution $u = \sin(x) + \cos(x)$, we get that
\begin{align*}
\int\frac{\sin(x)}{\sin(x) + \cos(x)}\mathrm{d}x = \frac{x}{2} - \frac{\ln(\sin(x) + \cos(x))}{2} + c
\end{align*}
Now it remains to apply the integration limits.
Hopefully this is helpful!
