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If I1 = ∫ln(sinx)dx for x ∈ [0, π/2] and I2 = ∫ln(sinx + cosx)dx for x [-π/4, π/4]

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If I1 = ∫ln(sinx)dx for x ∈ [0, π/2] and I2 = ∫ln(sinx + cosx)dx for x [-π/4, π/4] then find I1/I2.

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⇒ 2I2 = I1 ⇒ I1/I2 = 1/2

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