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Question: If \(I = \int _ { 0 } ^ { 100 \pi } \sqrt { ( 1 - \cos 2 x ) } d x\)then the value of \(I\) is...

If I=∫0100π(1−cos⁡2x)dxI = \int _ { 0 } ^ { 100 \pi } \sqrt { ( 1 - \cos 2 x ) } d xthen the value of II is

A

1002100 \sqrt { 2 }

B

2002200 \sqrt { 2 }

C

50250 \sqrt { 2 }

D

None of these

Answer

2002200 \sqrt { 2 }

Explanation

Solution

I=∫0π(1−cos⁡2x)dx+∫π2π(1−cos⁡2x)dx+…I = \int _ { 0 } ^ { \pi } \sqrt { ( 1 - \cos 2 x ) } d x + \int _ { \pi } ^ { 2 \pi } \sqrt { ( 1 - \cos 2 x ) } d x + \ldots

∵∫0naf(x)dx=n∫0af(x)dx\because \int _ { 0 } ^ { n a } f ( x ) d x = n \int _ { 0 } ^ { a } f ( x ) d x , if f(a+x)=f(x)f ( a + x ) = f ( x )

I=100∫0π(1−cos⁡2x)dxI = 100 \int _ { 0 } ^ { \pi } \sqrt { ( 1 - \cos 2 x ) } d x

I=1002∫0πsin⁡xdx=2002∫0π/2sin⁡xdxI = 100 \sqrt { 2 } \int _ { 0 } ^ { \pi } \sin x d x = 200 \sqrt { 2 } \int _ { 0 } ^ { \pi / 2 } \sin x d x

=2002[−cos⁡x]0π/2=2002= 200 \sqrt { 2 } [ - \cos x ] _ { 0 } ^ { \pi / 2 } = 200 \sqrt { 2 }.