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?
 A: Using
$$
\int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx
$$
Let 
\begin{align}
I &= \int_0^{2\pi} \frac{\sin x}{\left|\sin x\right|+\left|\cos x\right|} \,  dx \\
&= \int_0^{2\pi} \frac{\sin (2\pi -x)}{\left|\sin (2\pi -x)\right|+\left|\cos (2\pi -x)\right|} \, dx \\
&= \int_0^{2\pi} \frac{-\sin x}{\left|\sin x\right|+\left|\cos x\right|} \, dx 
\end{align}
$$
\Rightarrow \quad 2I = 0 \quad \Rightarrow \quad I =0
$$
Note: The rule is valid if $f(x)$ is continuous on $[0,a]$. Your function is continuous on $[0,2\pi]$. 
A: On splitting $\displaystyle\int_{0}^{2\pi}=\int_{0}^{\pi}+\int_{\pi}^{2\pi}$. For $\displaystyle\int_{\pi}^{2\pi}$, one can use $y=x-\pi$ and deduce that $\displaystyle\int_{\pi}^{2\pi}=-\int_{0}^{\pi}$.
A: $$\int_0^{2\pi}\frac{\sin x}{|\sin x| +|\cos x|}dx=$$
$$=\int_0^{\frac{\pi}{2}}\tfrac{\sin x}{\sin x +\cos x}dx+\int_{\frac{\pi}{2}}^{\pi}\tfrac{\sin x}{\sin x -\cos x}dx+\int_{\pi}^{\frac{3\pi}{2}}\tfrac{\sin x}{-\sin x-\cos x}+\int_{\frac{3\pi}{2}}^{2\pi}\tfrac{\sin x}{-\sin x +\cos x}dx=$$
$$=\int_0^{\frac{\pi}{2}}\tfrac{\sin x}{\sin x +\cos x}dx+\int_0^{\frac{\pi}{2}}\tfrac{\sin \left(x+\frac{\pi}{2}\right)}{\sin \left(x+\frac{\pi}{2}\right) -\cos \left(x+\frac{\pi}{2}\right) }dx+$$
$$+\int_0^{\frac{\pi}{2}}\tfrac{\sin(x+\pi)}{-\sin (x+\pi)-\cos(x+\pi)}dx+\int_0^{\frac{\pi}{2}}\tfrac{\sin\left( x+\frac{3\pi}{2}\right)}{-\sin\left( x+\frac{3\pi}{2}\right) +\cos\left( x+\frac{3\pi}{2}\right)}dx=$$
$$=\int_0^{\frac{\pi}{2}}\tfrac{\sin x}{\sin x +\cos x}dx+\int_0^{\frac{\pi}{2}}\tfrac{\cos{x}}{\cos{x}+\sin{x} }dx+$$
$$+\int_0^{\frac{\pi}{2}}\tfrac{-\sin{x}}{\sin{x}+\cos{x}}dx+\int_0^{\frac{\pi}{2}}\tfrac{-\cos{x}}{\cos{x} +\sin{x}}dx=0.$$
A: Here you can directly check $f(2\pi-x)=-f(x)$. So you need not really evaluate the integral. By this property of definite integrals (for continuous functions, obviously) you can directly conclude the integral to be $0$.
A: The point would be that $\displaystyle\int_0^\pi = -\int_\pi^{2\pi}.$
The function $x\mapsto \left| \sin x\right| + \left| \cos x\right|$ simply repeats on the interval $[\pi,2\pi]$ what it did on the interval $[0,\pi].$ It is a periodic function whose period is $\pi.$ The reason is that $x\mapsto \left|\sin x\right|$ and $x\mapsto\left|\cos x\right|$ separately just repeat on $[\pi,2\pi]$ what they did on $[0,\pi].$
But the function $x\mapsto\sin x$ repeats on $[\pi,2\pi]$ what it did on $[0,\pi]$ except that it gets multiplied $\text{by }-1.$ Or in other words $\sin(x+\pi) = -\sin x,$ and $\cos(x+\pi) = -\cos x.$
