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B
π4+log2
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C
π4−12log2
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D
π4−log2
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Solution
The correct option is Cπ4−12log2 ∫π20cosx1+cosx+sinxdx =∫π20cos2(x2)−sin2(x2)2cos2(x2)+2sin(x2)cos(x2)dx =12∫π201−tan2(x2)1+tan(x2)dx=12∫π20[1−tan(x2)]dx π4+log1√2=π4−12log2