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`        Integrate (sin8x-cos8x)/(1-2sin2xcos2x)`
7 years ago

147 Points
```										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
Please feel free to post as many doubts on our discussion forum as you can.If you find any question Difficult to understand - post it here and we will get you the answer and detailed  solution very  quickly. We are all IITians and here to help you in your IIT JEE preparation.All the best. Regards,Askiitians ExpertsBadiuddin
```
7 years ago
Anurag Kishore
37 Points
```										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
```
7 years ago
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