Why is $\int_{0}^{\pi}{1\over 1-\sin x}dx=2\int_{0}^{\pi\over 2}{1\over 1-\sin x}dx$? Why is $\int_{0}^{\pi}{1\over 1-\sin x}dx=2\int_{0}^{\pi\over 2}{1\over 1-\sin x}dx$, or to be accurate: why is $\int_{\pi\over 2}^{\pi}{1\over 1-\sin x}dx=\int_{0}^{\pi\over 2}{1\over 1-\sin x}dx$? 
At the very best, I know that the area $\sin x$ covers between $0$ to $\pi\over 2$ has the same magnitude between $\pi\over 2$ to $\pi$ but I fail to see how it formally leads to the identity aforementioned. Can you help me with understanding this?
 A: Make the substitution $u=\pi-x$; then
$$\int_{\pi/2}^\pi\frac{dx}{1-\sin x}=\int_{\pi/2}^0\frac{-du}{1-\sin u}=\int_0^{\pi/2}\frac{du}{1-\sin u}\;.$$
A: It appears by symmetry in that sin(x) x = 0 .. pi describes an arc that will define an area that is twice that of the half arc described by sin(x) x = 0 .. pi/2 so the second integral is half the area of the first hence multiplied by two makes them equal areas.
A: The integral of interest $\int_0^{\pi}\frac{dx}{1-\sin x}$ does not converge.  Rather, it diverges since 
$$1-\sin x=-\frac12 (x-\pi/2)^2+O\left((x-\pi/2)^4\right)$$
Even if we interpret the integral in the sense of a Cauchy Principal Value, then we have
$$\begin{align}
\text{P.V.}\int_0^{\pi}\frac{dx}{1-\sin x}&=\lim_{\epsilon \to 0}\left(\int_0^{\pi/2-\epsilon}\frac{dx}{1-\sin x}+\int_{\pi/2+\epsilon}^{\pi}\frac{dx}{1-\sin x}\right)\\\\
&=\lim_{\epsilon \to 0}\left(\int_0^{\pi/2-\epsilon}\frac{dx}{1-\sin x}+\int_{0}^{\pi/2-\epsilon}\frac{dx}{1-\sin x}\right)\\\\
&=2\lim_{\epsilon \to 0}\int_0^{\pi/2-\epsilon}\frac{dx}{1-\sin x}\\\\
&=\infty
\end{align}$$
So, although we can write
$$\int_{\pi/2+\epsilon}^{\pi}\frac{dx}{1-\sin x}=\int_{0}^{\pi/2-\epsilon}\frac{dx}{1-\sin x}$$
and therefore write
$$\int_0^{\pi/2-\epsilon}\frac{dx}{1-\sin x}+\int_{\pi/2+\epsilon}^{\pi}\frac{dx}{1-\sin x}=2\int_0^{\pi/2-\epsilon}\frac{dx}{1-\sin x}$$
the limit as $\epsilon$ goes to zero does not exist and we may not equate $\int_0^{\pi}\frac{dx}{1-\sin x}$ as asserted.
