Integrate 0 to π/2 of 2log(sinx)-log(sin2x) | `int_(0)^(pi//2) (2log sin x-log sin 2x)dx`

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The value of `int_(0)^(pi//2) (2log sin x-log sin 2x)dx`, isThe value of `int_(0)^(pi//2) (2log sin x-log sin 2x)dx`, is