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ksaimouli

  • one year ago

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  1. ksaimouli
    • one year ago
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    \[\frac{ -x^2 }{ 2 }\cos2x+\frac{ x }{ 2 }\sin2x+\frac{ 1 }{ 4 }\cos 2x\]

  2. ksaimouli
    • one year ago
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    to |dw:1360893059171:dw|

  3. ksaimouli
    • one year ago
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    i tried this i got \[\frac{ \pi^2 }{ 8 }-\frac{ 1 }{ 4 }\]

  4. hartnn
    • one year ago
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    how u got that ? show your work, i'll spot the error.

  5. ksaimouli
    • one year ago
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    oka

  6. ksaimouli
    • one year ago
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    |dw:1360893431147:dw|

  7. ksaimouli
    • one year ago
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    step 1 is that right

  8. hartnn
    • one year ago
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    yes.

  9. ksaimouli
    • one year ago
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    okay step 2|dw:1360893564991:dw|

  10. hartnn
    • one year ago
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    what about the lower limit of 0, will that come later ?

  11. hartnn
    • one year ago
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    so, you got pi^2/8-1/4 only by substituting upper limit ?

  12. hartnn
    • one year ago
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    thats correct by the way, but incomplete

  13. ksaimouli
    • one year ago
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    shut i forgot that literally i thought no value in polynomials but their is a value for trig

  14. ksaimouli
    • one year ago
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    ya i found the mistake thx

  15. hartnn
    • one year ago
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    yes, cos 0 is not 0

  16. hartnn
    • one year ago
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    welcome ^_^

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