Use the identity $\cos 2\theta  = 2{\cos ^2}\theta  - 1$ to prove that $\cos \frac{1}{2}x = \sqrt {\frac{{1 + \cos x}}{2}} ,{\text{ }}0 \leqslant x \leqslant \pi$.

Find a similar expression for $\sin \frac{1}{2}x,{\text{ }}0 \leqslant x \leqslant \pi$.

Hence find the value of $\int_0^{\frac{\pi }{2}} {\left( {\sqrt {1 + \cos x}  + \sqrt {1 - \cos x} } \right){\text{d}}x}$.

$\cos x = 2{\cos ^2}\frac{1}{2}x - 1$

$\cos \frac{1}{2}x =  \pm \sqrt {\frac{{1 + \cos x}}{2}}$     M1

positive as $0 \leqslant x \leqslant \pi$     R1

$\cos \frac{1}{2}x = \sqrt {\frac{{1 + \cos x}}{2}}$     AG

[2 marks]

$\cos 2\theta  = 1 - 2{\sin ^2}\theta$     (M1)

$\sin \frac{1}{2}x = \sqrt {\frac{{1 - \cos x}}{2}}$     A1

[2 marks]

$\sqrt 2 \int_0^{\frac{\pi }{2}} {\cos \frac{1}{2}x + \sin \frac{1}{2}x{\text{d}}x}$     A1

$= \sqrt 2 \left[ {2\sin \frac{1}{2}x - 2\cos \frac{1}{2}x} \right]_0^{\frac{\pi }{2}}$     A1

$= \sqrt 2 (0) - \sqrt 2 (0 - 2)$     A1

$= 2\sqrt 2$     A1

[4 marks]