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Question

The value of (sinxcosxcos2xcos4xcos8xcos16x)dx is equal to

A
sin16x1024+C
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B
cos32x1024+C
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C
cos32x1096+C
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D
cos32x1096+C
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Solution

The correct option is C cos32x1024+C
I=(sinxcosxcos2xcos4xcos8xcos16x)dx
I=12(2sinxcosx)(cos2xcos4xcos8xcos16x)dx
=12(sin2xcos2xcos4xcos8xcos16x)dx
=14(2sin2xcos2x)(cos4xcos8xcos16x)dx
=14(cos4xsin4x)(cos8xcos16x)dx
=18(sin8x)(cos8x)(cos16x)dx
=116sin16xcos16xdx
I=132 sin(32x)dx
put cos(32x)=t
(sin(32x))×32dx=dt
I=132×(132)dt
=1(32)2t+c
I=cos(32x)1024+c
sinxcosxcos2xcos4xcos8xcos16xdx=cos32x1024+c.

1118332_1195672_ans_3acfbc22ad1a45048785921dc855a5cd.jpeg

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