evaluate

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\[\frac{ -x^2 }{ 2 }\cos2x+\frac{ x }{ 2 }\sin2x+\frac{ 1 }{ 4 }\cos 2x\]
to |dw:1360893059171:dw|
i tried this i got \[\frac{ \pi^2 }{ 8 }-\frac{ 1 }{ 4 }\]

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how u got that ? show your work, i'll spot the error.
oka
|dw:1360893431147:dw|
step 1 is that right
yes.
okay step 2|dw:1360893564991:dw|
what about the lower limit of 0, will that come later ?
so, you got pi^2/8-1/4 only by substituting upper limit ?
thats correct by the way, but incomplete
shut i forgot that literally i thought no value in polynomials but their is a value for trig
ya i found the mistake thx
yes, cos 0 is not 0
welcome ^_^

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