Let $f\left(x\right)$ be a function satisfying $f\left(x\right)+f(\pi -x)=\frac{\pi }{4}$ for all $x\in R$ Then $\underset{0}{\overset{\pi }{\int }}f\left(x\right)\sin xdx=$

$I=\underset{0}{\overset{\pi }{\int }}f\left(x\right)\sin xdx=\underset{0}{\overset{\pi }{\int }}f(\pi -x)\sin xdx$

$\therefore I+I=\underset{0}{\overset{\pi }{\int }}\sin x[f\left(x\right)+f(\pi -x)]dx=\underset{0}{\overset{\pi }{\int }}\frac{\pi }{4}\sin xdx$

$\Rightarrow 2I=\frac{\pi }{4}\underset{0}{\overset{\pi }{\int }}\sin xdx=\frac{\pi }{4}\times 2\underset{0}{\overset{\pi }{\int }}\sin xdx=\frac{\pi }{2}\times 1\Rightarrow I=\frac{\pi }{4}$