ksaimouli
  • ksaimouli
evaluate
Mathematics
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.