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If $f$ is continuous function and $f(x+T)=f(x), \forall x \in R$, it is given that $\int_0^{N T} f(t) d t=N \int_0^T f(t) d t \cdot(N$ is natural number). Then, $\int_0^{50 \pi} \sqrt{1-\cos 2 x} d x=$
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$100 \sqrt{2}$
Let $\begin{aligned} I= & \int_0^{50 \pi} \sqrt{1-\cos 2 x} d x \\ = & 50 \int_0^\pi \sqrt{1-\cos 2 x} d x \\ & {[\because \sqrt{1-\cos 2(2 \pi+x)}=\sqrt{1-\cos 2 x}] } \\ I= & 50 \int_0^\pi \sqrt{2} \sin x d x \\ & =50 \sqrt{2}[-\cos x]_0^\pi \\ & =-50 \sqrt{2}[(-1)-1]=100 \sqrt{2}\end{aligned}$
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