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in Mathematics by (36.3k points)

∫(2 log sin x - log sin 2x) dx, for x ∈ [0, π/2].

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Let I = ∫(2 log sin x - log sin 2x) dx, for x ∈ [0, π/2]

= ∫{2 log sin x - log(2sin x.cos x)} dx, for x ∈ [0, π/2]

= ∫{2 log sin x - log2 - log sin x - log cos x} dx, for x ∈ [0, π/2]

= ∫(log sin x - log2 - log cos x) dx, for x ∈ [0, π/2]

= ∫(log sin x dx - ∫log2 dx - ∫log cos x dx), for x ∈ [0, π/2]

= ∫{(log sin x dx - log2∫dx - ∫log cos((π/2) - x)dx}, for x ∈ [0, π/2]

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