Answer
\[\frac{1}{2}\sqrt{2}\]
Work Step by Step
Considering indefinite integral
\[I=\int\sqrt{1+\cos 2x}\:dx\]
\[\left[\cos 2x=2\cos^2 x-1\right]\]
\[I=\int\sqrt{2\cos^2 x}\:dx\]
\[I=\sqrt{2}\int\cos x\:dx\]
$\;\;\; \;\;\;\;\;\;\;\;\;\;I=\sqrt{2}\sin x$ ____(1)
Let \[I_1=\int^{\frac{π}{6}}_{0}\sqrt{1+\cos 2x}\:dx\]
Using (1)
\[I_1=\left[\sqrt{2}\sin x\right]^{\frac{π}{6}}_{0}\]
\[I_1=\sqrt{2}\left[\sin\frac{π}{6}-\sin(0)\right]\]
\[I_1=\frac{1}{2}\sqrt{2}\]
Hence $I_1=\frac{1}{2}\sqrt{2}$.
