Answer
\[ = \sqrt 2 \]
Work Step by Step
\[\begin{gathered}
\int\limits_0^{\frac{\pi }{2}} {\sqrt {1 + \cos 2x} \,\,dx} \hfill \\
\hfill \\
using\,\,the\,\,identity\,\,\,1 = {\sin ^2}x + {\cos ^2}x\,\,and\,\,\cos \,2x = {\cos ^2}x - {\sin ^2}x \hfill \\
integral\,becomes \hfill \\
\hfill \\
\int\limits_0^{\frac{\pi }{2}} {\sqrt {1 + \cos 2x} \,\,dx} = \int\limits_0^{\frac{\pi }{2}} {\sqrt {{{\cos }^2}x + {{\sin }^2}x + {{\cos }^2}x - {{\sin }^2}x} \,dx} \hfill \\
\hfill \\
simplify\,\,inside\,\,the\,\,\,radical \hfill \\
\hfill \\
= \sqrt 2 \int\limits_0^{\frac{\pi }{2}} {\cos xdx} \hfill \\
\hfill \\
integrate\,and\,use\,\,the\,\,ftc \hfill \\
\hfill \\
= \sqrt 2 \,\,\left[ {\sin x} \right]_0^{\frac{\pi }{2}} \hfill \\
\hfill \\
= \sqrt 2 \,\left( {1 - 0} \right) = \sqrt 2 \hfill \\
\end{gathered} \]
