$$\int_0^{100\pi } {\sqrt {1 + \cos \,2x} } \,dx$$     is equal to :

Solution :
$$I = \int_0^{100\pi } {\sqrt 2 } \left| {\cos \,x} \right|dx.$$     But $$\left| {\cos \,x} \right|$$   is a periodic function of period $$\pi .$$
$$\eqalign{ & \therefore I = 100\int_0^\pi {\sqrt 2 \left| {\cos \,x} \right|dx} \cr & = 100\sqrt 2 \left\{ {\int_0^{\frac{\pi }{2}} {\left| {\cos \,x} \right|dx + } \int_{\frac{\pi }{2}}^\pi {\left| {\cos \,x} \right|dx} } \right\} \cr & = 100\sqrt 2 \left\{ {\int_0^{\frac{\pi }{2}} {\cos \,x\,dx + } \int_{\frac{\pi }{2}}^\pi { - \cos \,x\,dx} } \right\} \cr & = 100\sqrt 2 \left\{ {\left[ {\sin \,x} \right]_0^{\frac{\pi }{2}} + \left[ { - \sin \,x} \right]_{\frac{\pi }{2}}^x} \right\} \cr & = 100\sqrt 2 \left\{ {1 - 0 - \left( {0 - 1} \right)} \right\} \cr & = 200\sqrt 2 \cr} $$