Question

The value of ${\int }_0^{100\pi }\sqrt{1-\cos 2x}dx$ is equal to?

• A. $100\sqrt{2}$
• B. $50\sqrt{2}$
• C. $200\sqrt{2}$
• D. $400\sqrt{2}$

Answer 1

The correct option is C $200\sqrt{2}$

Let $I={\int }_0^{100\pi }\sqrt{1-\cos 2x}dx=\sqrt{2}{\int }_0^{100\pi }\left|\sin x\right|dx$
Since $\left|\sin x\right|$ is a periodic function of the period $\pi$
$\therefore 100\sqrt{2}{\int }_0^{\pi }\left|\sin x\right|dx=100\sqrt{2}{\int }_0^{\pi }\left|\sin x\right|dx$ $[\because \sin x\ge 0$ in $[0,\pi ]]$
$I=100\sqrt{2}{[-\cos x]}_0^{\pi }=200\sqrt{2}$