Question

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

• A. $150\sqrt{2}$
• B. $100\sqrt{2}$
• C. $200\sqrt{2}$
• D. $50\sqrt{2}$

Answer 1

Answer: (C)

$200\sqrt{2}$

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

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

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

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

Now we know that if $T$ is the time period of $F\left(x\right)$, then

${\int }_0^{nT}F\left(x\right)dx=n{\int }_0^{T}F\left(x\right)dx$

And the time period of $|\sin x|$ has period of $\pi$.

Hence, the given integration becomes-

$100\sqrt{2}{\int }_0^{\pi }|\sin x|dx$ and in the interval $[0,\pi ]$,

$\sin x$ is positive, and hence $|\sin x|=\sin x.$

Thus integration becomes $100\sqrt{2}{\int }_0^{\pi }\sin xdx$

$=100\sqrt{2}{[-\cos ]}_0^{\pi }$

$=100\sqrt{2}\times 2=200\sqrt{2}$