Integrals of Trigonometry functions Show that
$$\int^\pi_0 \frac{\sqrt{1+\cos x}}{\sqrt{1+\cos x}+\sqrt{1-\cos x}}\,dx =\int^\pi_0 \frac{\sqrt{1-\cos x}}{\sqrt{1+\cos x}+\sqrt{1-\cos x}}\,dx$$
I tried to use to check if
$$\int^\pi_0\left(\frac{\sqrt{1+\cos x}}{\sqrt{1+\cos x}+\sqrt{1-\cos x}} - \frac{\sqrt{1-\cos x}}{\sqrt{1+\cos x}+\sqrt{1-\cos x}}\right)\,dx=0$$
but it didn't turn out well. 
Please help.
 A: Your idea actually works; you just have to pursue it.  Abbreviating $\cos x$ to $c$ (and $\sin x$ to $s$), we have
$${\sqrt{1+c}-\sqrt{1-c}\over\sqrt{1+c}+\sqrt{1-c}}={(\sqrt{1+c}-\sqrt{1-c})^2\over(1+c)-(1-c)}={1+c-2\sqrt{1-c^2}+1-c\over2c}={1-s\over c}={c\over1+s}$$
(NB: $\sqrt{1-c^2}=s$ has the correct sign, since $s=\sin x\ge0$ for $0\le x\le\pi$.)  The substitution $u=1+\sin x$, $du=\cos x\ dx$ gives
$$\int_0^\pi{\sqrt{1+\cos x}-\sqrt{1-\cos x}\over\sqrt{1+\cos x}+\sqrt{1-\cos x}}dx=\int_0^\pi{\cos x\ dx\over1+\sin x}=\int_1^1{du\over u}=0$$
A: HINT:
Use $$\int_a^bf(x)\ dx=\int_a^bf(a+b-x)\ dx$$  which can be derived using  $a+b-x=y$
and $\cos(\pi-x)=-\cos x$
A: HINT: Substitute $u=\pi-x$ in the first integral.
A: The point would be that the integral of the first function from $0$ to $\pi/2$ is the same as the integral of the second function from $\pi/2$ to $\pi$ and vice-versa.
\begin{align}
& \int_{x=0}^{x=\pi/2} \Big(\text{some function of } \cos x\Big)\,dx \\[6pt]
= {} & \int_{u=\pi}^{u=\pi/2} \Big(\text{the same function of }\cos (\pi-u)\Big)\,(-du) \\[6pt]
= {} & \int_{u=\pi}^{u=\pi/2} \Big(\text{the same function of }(-\cos u)\Big)\,(-du) \\[6pt]
= {} & \int_{\pi/2}^\pi \Big(\text{the same function of }(-\cos u)\Big)\,du \\[6pt]
= {} & \int_{\pi/2}^\pi \Big(\text{the same function of }(-\cos x)\Big)\,dx
\end{align}
A: Hint:
$$\frac{\sqrt{1+\cos x}}{\sqrt{1+\cos x}+\sqrt{1-\cos x}}=\frac1{2\cos x}+\frac12-\frac12\tan x$$
