Practicing Success
The value of the integral $\int\limits_0^{400 \pi} \sqrt{1-\cos 2 x} d x$, is |
$200 \sqrt{2}$ $400 \sqrt{2}$ $800 \sqrt{2}$ none of these |
$800 \sqrt{2}$ |
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] $\Rightarrow I=400 \sqrt{2} \int\limits_0^\pi \sin x d x=400 \sqrt{2}[-\cos x]_0^\pi=800 \sqrt{2}$ |