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Evaluate ∫((sinx + cosx)/(9 + 16sin2x))dx for x ∈ [0, π/4]

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Evaluate ∫((sinx + cosx)/(9 + 16sin2x))dx for x ∈ [0, π/4]

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Best answer

Put y = sinx − cosx

dy = (cosx + sinx)dx 

y2 = sin2x + cos2x − 2sinxcosx = 1 − sin2x

Therefore, sin 2x = 1− y2

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