Easiest way to show that $\int_0^{2\pi}\frac{\sin x\,dx}{\left|\sin x\right| +\left|\cos x\right|}=0$ What's the easiest way to show that $$\int_0^{2\pi}\frac{\sin x \,dx}{\left|\sin x\right| +\left|\cos x\right|}=0$$
I was thinking about to change the interval of integration and to show that the function is odd.
$\displaystyle y=x-\pi \Rightarrow \int_{-\pi}^\pi \frac{-\sin y \, dy}{\left|\sin y\right| + \left|\cos y\right|}=0$ (I guess). Does it seem legit?

$$\int_0^{2\pi}\frac{\sin x}{|\sin x| +|\cos x|}dx=$$
$$=\int_0^{\frac{\pi}{2}}\tfrac{\sin x}{\sin x +\cos x}dx+\int_{\frac{\pi}{2}}^{\pi}\tfrac{\sin x}{\sin x -\cos x}dx+\int_{\pi}^{\frac{3\pi}{2}}\tfrac{\sin x}{-\sin x-\cos x}+\int_{\frac{3\pi}{2}}^{2\pi}\tfrac{\sin x}{-\sin x +\cos x}dx=$$
$$=\int_0^{\frac{\pi}{2}}\tfrac{\sin x}{\sin x +\cos x}dx+\int_0^{\frac{\pi}{2}}\tfrac{\sin \left(x+\frac{\pi}{2}\right)}{\sin \left(x+\frac{\pi}{2}\right) -\cos \left(x+\frac{\pi}{2}\right) }dx+$$
$$+\int_0^{\frac{\pi}{2}}\tfrac{\sin(x+\pi)}{-\sin (x+\pi)-\cos(x+\pi)}dx+\int_0^{\frac{\pi}{2}}\tfrac{\sin\left( x+\frac{3\pi}{2}\right)}{-\sin\left( x+\frac{3\pi}{2}\right) +\cos\left( x+\frac{3\pi}{2}\right)}dx=$$
$$=\int_0^{\frac{\pi}{2}}\tfrac{\sin x}{\sin x +\cos x}dx+\int_0^{\frac{\pi}{2}}\tfrac{\cos{x}}{\cos{x}+\sin{x} }dx+$$
$$+\int_0^{\frac{\pi}{2}}\tfrac{-\sin{x}}{\sin{x}+\cos{x}}dx+\int_0^{\frac{\pi}{2}}\tfrac{-\cos{x}}{\cos{x} +\sin{x}}dx=0.$$
