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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Badiuddin

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