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Integrate (sin 8 x-cos 8 x)/(1-2sin 2 xcos 2 x)

Integrate (sin8x-cos8x)/(1-2sin2xcos2x)

Grade:12

2 Answers

Badiuddin askIITians.ismu Expert
148 Points
12 years ago

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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Anurag Kishore
37 Points
12 years ago

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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