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Question

-π/2π/2loga-sin θa+sin θ dθ

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Solution

Let, I=-π2π2loga-sinθa+sinθdθHere, fθ=loga-sinθa+sinθConsider, f-θ=loga-sin-θa+sin-θ=-loga-sinθa+sinθ=-fθi.e., fθ is odd function.Therefore, I=0

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