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Integrate (sin8x-cos8x)/(1-2sin2xcos2x)
Dear Mithresh
∫(sin8x-cos8x)/(1-2sin2xcos2x) dx
= ∫2(sin4x+cos4x)(sin2x-cos2x)/(2-sin22x) dx
=-1/2∫[(1-cos2x)2+(1+cos2x)2](cos2x)/(2-sin22x) dx
=-∫[1+cos22x](cos2x)/(2-sin22x) dx
=-∫(cos2x) dx
=-1/2 sin2x + c
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Dear,
(sin8x-cos8x)/(1-2sin2xcos2x).dx
sin8x - cos8x = (sin4x - cos4x)(sin4x+cos4x)
= (sin2x+cos2x)(sin2x-cos2x)[(sin2x+cos2x)2 - 2sin2x cos2x]
= (sin2x - cos2x)(1-2sin2x cos2x)
Put the value in integration,
(sin2x-cos2x). dx
= -cos2x . dx
= (-sin2x)/2
answer may vary which can be obtained by transformation.
Thanks
Anurag
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