The equation ${\int }_{-\pi /4}^{\pi /4} \left\{a|\sin x|+\frac{b\sin x}{1+\cos x}+c\right\}dx=0,$ where $a, b, c$ are constants, gives a relation between

We have,

$\begin{array}{l}\left.{\int }_{-\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.}^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.} a|\sin x|+\frac{b\sin x}{1+\cos x}+c\right\}dx=0\\ \Rightarrow a{\int }_{-\pi }^{\pi /4} |\sin x|dx+b{\int }_{-\pi /4}^{\pi /4}\frac{\sin x}{1+\cos x}dx+c{\int }_{-\pi /4}^{\pi /4} dx=0\\ \Rightarrow 2a{\int }_0^{\pi } |\sin x|dx+b\times 0+2c{\int }_0^{\pi } dx=0\\ \Rightarrow 2a{\int }_0^{\pi } \sin xdx+2c\times \frac{\pi }{4}=0\\ -2a\left(\cos \frac{\pi }{4}-\cos 0\right)+\frac{\pi c}{2}=0\\ -2a\left(\frac{1}{\sqrt{2}}-1\right)+\frac{\pi c}{2}=0\Rightarrow a(2-\sqrt{2})+\frac{\pi c}{2}=0\end{array}$