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Question

The value of π0sin4xdx is:

A
3π16
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B
316
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C
0
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D
3π8
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Solution

The correct option is D 3π8
sin4(x)=sin2x.sin2x=sin2x(1cos2x)
=sin2xsin2xcos2x
=1cos2x24sin2x.cos2x4
=1cos2x2sin22x4
=1cos2x21cos4x8
=1218+cos4x8cos2x2
=616+cos4x8cos2x2
=38+cos4x8cos2x2
Therefore
π0sin4xdx=π0(38+cos4x8cos2x2)dx
=[3x8+sin4x32sin2x4]π0
=3π8.

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