    Question

# Assertion :The value of ∫π/20sin6xdx=5π16 Reason: If n is even, then ∫π/20sinnxdx equals n−1nn−3n−2n−5n−4...12×π2

A
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
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B
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
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C
Assertion is correct but Reason is incorrect
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D
Assertion is incorrect but Reason is correct.
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Solution

## The correct option is D Assertion is incorrect but Reason is correct.I=∫π20sin6(x).dx=∫π02sin6(π2−x).dx=∫π20cos6(x).dxHence2I=∫π20sin6(x)+cos6(x).dxsin6(x)+cos6(x)=[(sin2(x)+cos2(x))(sin4(x)+cos4(x)−sin2(x).cos2(x))]=sin4(x)+cos4(x)−sin2(x).cos2(x))=(sin2(x)+cos2)2−2sin2x.cos2x−sin2(x)cos2(x)=1−3sin2x.cos2x=1−3sin22(x)4=1−3(1−cos4x)8=58+3cos4x8Hence2I=∫π2058+3cos4x8.dx2I=5π16I=5π32.Hence Assertion is false.  Suggest Corrections  0      Similar questions  Related Videos   Arithmetic Progression - Sum of n Terms
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