Tính \(\int\limits_0^\pi {\sqrt {1 + \sin 2x} dx} \).

Ta có \(\int\limits_0^\pi {\sqrt {1 + \sin 2x} dx} = \int\limits_0^\pi {\sqrt {{{\sin }^2}x + {{\cos }^2}x + 2\sin x.\cos x} dx} \).

\( = \int\limits_0^\pi {\sqrt {{{\left( {\sin x + \cos x} \right)}^2}} dx} = \int\limits_0^\pi {\left| {\sin x + \cos x} \right|dx} \).

\[ = \int\limits_0^\pi {\left| {\sqrt 2 \cos \left( {x - \frac{\pi }{4}} \right)} \right|dx} = \sqrt 2 \int\limits_0^{\frac{{3\pi }}{4}} {\cos \left( {x - \frac{\pi }{4}} \right)dx} - \sqrt 2 \int\limits_{\frac{{3\pi }}{4}}^\pi {\cos \left( {x - \frac{\pi }{4}} \right)dx} \].

\[ = \sqrt 2 \sin \left. {\left( {x - \frac{\pi }{4}} \right)} \right|_0^{\frac{{3\pi }}{4}} - \sqrt 2 \sin \left. {\left( {x - \frac{\pi }{4}} \right)} \right|_{\frac{{3\pi }}{4}}^\pi = \sqrt 2 + 1 - \left( {1 - \sqrt 2 } \right) = 2\sqrt 2 \].

Vậy \(\int\limits_0^\pi {\sqrt {1 + \sin 2x} dx} = 2\sqrt 2 \).