A community for students.
Here's the question you clicked on:
 0 viewing
ksaimouli
 3 years ago
evaluate
ksaimouli
 3 years ago
evaluate

This Question is Closed

ksaimouli
 3 years ago
Best ResponseYou've already chosen the best response.0\[\frac{ x^2 }{ 2 }\cos2x+\frac{ x }{ 2 }\sin2x+\frac{ 1 }{ 4 }\cos 2x\]

ksaimouli
 3 years ago
Best ResponseYou've already chosen the best response.0to dw:1360893059171:dw

ksaimouli
 3 years ago
Best ResponseYou've already chosen the best response.0i tried this i got \[\frac{ \pi^2 }{ 8 }\frac{ 1 }{ 4 }\]

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.1how u got that ? show your work, i'll spot the error.

ksaimouli
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1360893431147:dw

ksaimouli
 3 years ago
Best ResponseYou've already chosen the best response.0okay step 2dw:1360893564991:dw

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.1what about the lower limit of 0, will that come later ?

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.1so, you got pi^2/81/4 only by substituting upper limit ?

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.1thats correct by the way, but incomplete

ksaimouli
 3 years ago
Best ResponseYou've already chosen the best response.0shut i forgot that literally i thought no value in polynomials but their is a value for trig

ksaimouli
 3 years ago
Best ResponseYou've already chosen the best response.0ya i found the mistake thx
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.