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∫ sin x + cos x …

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∫ sin x + cos x DX by under root sin 2x
  • 2 answers
Sin-1(sinx-cosx)

O Singh 7 years ago

Put sin2x=1-(sinx-cosx)^2 and then substitute sinx - cos x as t . U will get √ 1-t^2 in denominator and then solve it.
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