Question

The value of $\underset{0}{\overset{100\pi }{\int }}\sqrt{1-\cos 2x}dx$ is

• A. $200\sqrt{2}$
• B. $400$
• C. $100\sqrt{2}$
• D. $0$

Answer 1

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

Let $I=\underset{0}{\overset{100\pi }{\int }}\sqrt{1-\cos 2x}dx$
$\Rightarrow I=\underset{0}{\overset{100\pi }{\int }}\sqrt{2{\sin }^{2}x}dx=\sqrt{2}\underset{0}{\overset{100\pi }{\int }}|\sin x|dx$
Using the property:
$\underset{a}{\overset{a+nT}{\int }}f\left(x\right)dx=n\underset{0}{\overset{T}{\int }}f\left(x\right)dx\left[T\text{is the period of}f\right(x\left)\right]$

Here, the period of $|\sin x|$ is $\pi$, so
$I=100\sqrt{2}\underset{0}{\overset{\pi }{\int }}|\sin x|dx=100\sqrt{2}\underset{0}{\overset{\pi }{\int }}\sin xdx=100\sqrt{2}(-\cos x{)}_0^{\pi }\therefore I=200\sqrt{2}$