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The value of 0sin2xsin1tdt+0cos2xcos1tdt is

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a
π2
b
1
c
π4
d
None of these

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detailed solution

Correct option is C

f(x)=∫0sin2⁡x sin−1⁡tdt+∫0cos2⁡x cos−1⁡tdtf′(x)=sin−1⁡sin2⁡x2sin⁡xcos⁡x+cos−1⁡cos2⁡x(−2cos⁡xsin⁡x)=sin⁡2xsin−1⁡(sin⁡x)−cos−1⁡(cos⁡x)=sin⁡2x(x−x)=0f (x) = constantFind f (x) for any value of x∫01/2 π2dx=π2⋅12=π4


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