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log4x=1-8sin^(2)x cos^(2)x...

log4x=1-8sin^(2)x cos^(2)x

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cos4x=1-8sin^(2)x cos^(2)x

cos4x=1-8sin^(2)x cos^(2)x

cos4x=1-8sin^(2)x cos^(2) x

Prove that :cos4x=1-8sin^(2)x cos^(2)x

The minimum value of 1-8sin^(2) x cos^(2) x is

int (sin^(8)x-cos^(8)x)/(1-2sin^(2)x cos^(2)x)dx =

int(sin^(8)x-cos^(8)x)/(1-2sin^(2)x cos^(2)x)dx=

int(sin^(8)x-cos^(8)x)/(1-2sin^(2)x cos^(2)x)dx

If int(sin^(8)x-cos^(8)x)/(1-2sin^(2)x cos^(2)x)dx=a sin2x+C then a=