How to integrate $\int_0^{\pi/2} \ \frac{\cos{x}}{\sqrt{1+\cos{x}}} \, \mathrm{d}x.$ I need to somehow evaluate the following:
$$ \int_0^{\pi/2} \ \dfrac{\cos{x}}{\sqrt{1+\cos{x}}} \, \mathrm{d}x. $$
Can anyone give me any hints/pointers? I've tried to use parts, and some feeble substitutions, but to no avail :( 
Thanks
 A: Write $\cos x = 2\cos^2\left(\frac{x}2\right) -1$. Then $\sqrt{1+\cos x} = \sqrt 2 \cos\left(\frac x 2\right)$ when $x\in[0,\pi/2]$.
So we are seeking $$\frac{1}{\sqrt 2}\int_0^{\frac \pi 2}\frac{2\cos^2\left(\frac{x}2\right) -1}{\cos\left(\frac x 2\right)}\,dx$$
Letting $u=x/2$ then $dx=2du$ and we get the problem:
$$\sqrt 2\int_0^{\frac\pi 4} (2\cos u-\sec u)\,du$$
Integral of $\sec$ is described here. 
A: Write denominator in form of $\cos{(x/2)}$ and then substitute $\pi/2-x=t$.
More hints : 

$\sin x=2\sin(x/2)\cos(x/2)$

A: Hint: $$\frac{1+\cos x}{2} = \cos^2\left(\frac x2\right)\Rightarrow\\\int\frac{\cos x}{\sqrt{1+\cos x}} \, \mathrm{d}x=\\\int\frac{\cos x}{\sqrt{2}\cos\left(\frac x2\right)} \, \mathrm{d}x=\\\int \frac{2\cos^2\left(\frac x2\right)-1}{\sqrt{2}\cos\left(\frac x2\right)} \, \mathrm{d}x=\\\int \sqrt{2}\cos\left(\frac x2\right) \, \mathrm{d}x-\frac{1}{\sqrt{2}\cos\left(\frac x2\right)} \, \mathrm{d}x=\\\sqrt{2}\int \cos\left(\frac x2\right) \, \mathrm{d}x- \int \frac{1}{\sqrt{2}\cos\left(\frac x2\right)} \, \mathrm{d}x=\\2\sqrt{2}\sin\left(\frac x2\right)-\frac{1}{\sqrt2}\int \sec\left(\frac x2\right) \, \mathrm{d}x=\\2\sqrt{2}\sin\left(\frac x2\right)-\frac{1}{\sqrt2}\cdot 2 \left(-\ln\left(\cos\left(\frac x4\right)-\sin\left(\frac x4\right)\right)+\ln\left(\cos\left(\frac x4\right)+\sin\left(\frac x4\right)\right)\right)+C=\\\sqrt{2}\left(\sqrt2\sin\left(\frac x2\right)+\ln\left(\cos\left(\frac x4\right)-\sin\left(\frac x4\right)\right)-\ln\left(\cos\left(\frac x4\right)+\sin\left(\frac x4\right)\right) \right)+C$$
Thus $$\int_0^{\pi/2} \ \dfrac{\cos{x}}{\sqrt{1+\cos{x}}} \, \mathrm{d}x=\\\sqrt{2}\left(\sqrt2\sin\left(\frac \pi4\right)+\ln\left(\cos\left(\frac \pi8\right)-\sin\left(\frac \pi8\right)\right)-\ln\left(\cos\left(\frac \pi8\right)+\sin\left(\frac \pi8\right)\right) \right)-\sqrt{2}\left(\sin\left(0\right)+\ln\left(\cos\left(0\right)-\sin\left(0\right)\right)-\ln\left(\cos\left(0\right)+\sin\left(0\right)\right)\right)=\\2-2 \sqrt2 \cdot \frac{1}{\tanh\left(\tan(\frac \pi8)\right)}\approx 0,75$$
