The value of the integral ∫limits_0^400 π √(1-cos 2 x) d x, is

We have,

$I=\int\limits_0^{400 \pi} \sqrt{1-\cos 2 x}=\int\limits_0^{400 \pi} \sqrt{2}|\sin x| d x$

$\Rightarrow I=\sqrt{2} \times 400 \int\limits_0^\pi|\sin x| d x$           [∵ |sin x| is periodic with period π]

$\Rightarrow I=400 \sqrt{2} \int\limits_0^\pi \sin x d x=400 \sqrt{2}[-\cos x]_0^\pi=800 \sqrt{2}$