Question

Let $I={\int }_0^{100\pi }\sqrt{(1-\cos 2x)}dx$, then

• A. $I=0$
• B. $I=200\sqrt{2}$
• C. $I=\pi \sqrt{2}$
• D. $I=100$

Answer 1

The correct option is B $I=200\sqrt{2}$

$I={\int }_0^{100\pi }\sqrt{1-\cos 2x}dx$

$={\int }_0^{100\pi }\sqrt{2{\sin }^{2}x}dx$

$=\sqrt{2}{\int }_0^{100\pi }\sin xdx$

$=100\sqrt{2}{\int }_0^{\pi }|\sin x|dx$

$=100\sqrt{2}|\cos x{|}_0^{\pi }dx$

$=200\sqrt{2}$