zhanabe8lGamartinar zhanabe8lGamartinar
  • 26-11-2016
  • Mathematics
contestada

How do you take the antiderivative of (sinxcosx)^2?

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nobillionaireNobley
nobillionaireNobley nobillionaireNobley
  • 02-12-2016
∫(sinx cosx)^2dx = ∫(1/2sin 2x)^2 dx = 1/4∫sin^2 2x dx = 1/4∫1/2(1 - cos 4x)dx = 1/8∫(1 - cos 4x) dx = 1/8[x - sin 4x / 4] + c = 1/32(4x - sin 4x) + c
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