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The value of the integral 0100π1cos 2xdx is

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a
1002
b
2002
c
0
d
100π

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detailed solution

Correct option is B

We have 1−cos⁡ 2x=2|sin⁡ x|. Since the period of |sin x| is π, so ∫0100π 1−cos⁡ 2xdx=2∫0100π  |sin⁡ x|dx=1002∫0π sin⁡ x dx=2002.


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