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Question

π20sin2xsin2x1+sin8x dx is equal to

A
1
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B
π2
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C
π4
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D
π8
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Solution

The correct option is D π8

π202sinxcosxsin2x1+sin8x dx=π202sin3xcosx1+sin8x dx
Let sin4x=t 4sin3xcosx=dt
=1012dt1+t2=12tan1t10=12tan110=π8

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