Quantcast

Got Homework?

Connect with other students for help. It's a free community.

  • across
    MIT Grad Student
    Online now
  • laura*
    Helped 1,000 students
    Online now
  • Hero
    College Math Guru
    Online now

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

UnkleRhaukus

\[\frac23\int\limits_{0}^{3}\cos(x)\sin\left(\frac{\pi n x}{3/2}\right)\text dx\]

  • one year ago
  • one year ago

  • This Question is Closed
  1. AravindG
    Best Response
    You've already chosen the best response.
    Medals 0

    @UnkleRhaukus whats in ur mind ?

    • one year ago
  2. UnkleRhaukus
    Best Response
    You've already chosen the best response.
    Medals 2

    i tried that , but i keep getting the wrong answer

    • one year ago
  3. UnkleRhaukus
    Best Response
    You've already chosen the best response.
    Medals 2

    \[\begin{align*} % b_n b_n&=\frac23\int\limits_{0}^{3}\cos(x)\sin\left(\frac{\pi n x}{3/2}\right)\text dx\\ &=\frac13\int\limits_{0}^{3}\sin\left((\tfrac23\pi n+1)x\right)-\sin\left((\tfrac23\pi n-1)x\right)\text dx\\\ &=-\frac13\left.\left(\frac{\cos\left((\tfrac23\pi n+1)x\right)}{\tfrac23\pi n+1}-\frac{\cos\left((\tfrac23\pi n-1)x\right)}{\tfrac23\pi n-1}\right)\right|_0^3\\ &=\frac{1-\cos\left(2\pi n+3\right)}{2\pi n+3}-\frac{1-\cos\left(2\pi n-3\right)}{2\pi n-3}\\ &=\frac{1-\cos\left(3\right)}{2\pi n+3}-\frac{1-\cos\left(3\right)}{2\pi n-3}\\ &=\left(1-\cos\left(3\right)\right)\left(\frac{1}{2\pi n+3}-\frac{1}{2\pi n-3}\right)\\ &=\left(1-\cos\left(3\right)\right)\left(\frac{2\pi n-3}{(2\pi n+3)(2\pi n-3)}-\frac{2\pi n+3}{(2\pi n-3)(2\pi n+3)}\right)\\ &=\left(1-\cos\left(3\right)\right)\left(\frac{-6}{4\pi^2n^2-9}\right)\\ \end{align*}\]

    • one year ago
  4. AravindG
    Best Response
    You've already chosen the best response.
    Medals 0

    funny my comment gt deleted !! anyways

    • one year ago
  5. UnkleRhaukus
    Best Response
    You've already chosen the best response.
    Medals 2

    am i using the wrong trig substitution?

    • one year ago
  6. UnkleRhaukus
    Best Response
    You've already chosen the best response.
    Medals 2

    @scarydoor

    • one year ago
  7. scarydoor
    Best Response
    You've already chosen the best response.
    Medals 2

    sorry I'm not very good at the trigonometric identities... I'm a bit lost in your derivation, due to my own limits.

    • one year ago
  8. UnkleRhaukus
    Best Response
    You've already chosen the best response.
    Medals 2

    thanks for having a look,

    • one year ago
  9. scarydoor
    Best Response
    You've already chosen the best response.
    Medals 2

    http://www.sosmath.com/trig/Trig5/trig5/trig5.html Are you using the second to last 'product to sum' formula here?

    • one year ago
  10. UnkleRhaukus
    Best Response
    You've already chosen the best response.
    Medals 2

    i used the last product to sum formula, would the result be different if i used the second to last formula/?

    • one year ago
  11. scarydoor
    Best Response
    You've already chosen the best response.
    Medals 2

    not sure. I might have just made a mistake in thikning it was the second to last one. The second to last one might not even apply properly.

    • one year ago
  12. AccessDenied
    Best Response
    You've already chosen the best response.
    Medals 0

    I think they'd both be the 'same,' except one was modified slightly with a negative factored out of sine to get it into a nicer form. The multiplication 'should' be commutative there...

    • one year ago
  13. scarydoor
    Best Response
    You've already chosen the best response.
    Medals 2

    On the second line, the second term in the integral you've got: \[-sin[(\frac{2}{3} \pi n - 1)x] \] whereas I've got it as: \[-sin[(1-\frac{2}{3}\pi n) x] \] Maybe you did something else to get your expression that I haven't seen. Or else maybe that's an error?

    • one year ago
  14. UnkleRhaukus
    Best Response
    You've already chosen the best response.
    Medals 2

    ah, i think thats it.

    • one year ago
  15. scarydoor
    Best Response
    You've already chosen the best response.
    Medals 2

    Wolfram Alpha thinks the solution is not nice looking: http://www.wolframalpha.com/input/?i=integral+of+1%2F3+*+cos+x+sin%28%282pi+n+x+%29+%2F+3%29 (can be useful to confirm that it's supposed to look pretty bad...)

    • one year ago
  16. scarydoor
    Best Response
    You've already chosen the best response.
    Medals 2

    I just typed in the indefinite integral form, to see how messy the anti-derivative is. Then you would need to evaluate at x=0 and x=3 in the usual way.

    • one year ago
  17. scarydoor
    Best Response
    You've already chosen the best response.
    Medals 2

    I need to sleep. good luck with it.

    • one year ago
  18. UnkleRhaukus
    Best Response
    You've already chosen the best response.
    Medals 2

    thank you

    • one year ago
  19. UnkleRhaukus
    Best Response
    You've already chosen the best response.
    Medals 2

    wolfram is having trouble because it dosent realize n in an integer but ive got what i wanted now i have the negative sign fixed thanks again @scarydoor \[\begin{align*} % b_n b_n&=\frac23\int\limits_{0}^{3}\cos(x)\sin\left(\frac{\pi n x}{3/2}\right)\text dx\\ &=\frac13\int\limits_{0}^{3}\sin\left((1+\tfrac23\pi n)x\right)-\sin\left((1-\tfrac23\pi n)x\right)\text dx\\\ &=-\frac13\left.\left(\frac{\cos\left((1+\tfrac23\pi n)x\right)}{1+\tfrac23\pi n}-\frac{\cos\left((1-\tfrac23\pi n)x\right)}{1-\tfrac23\pi n}\right)\right|_0^3\\ &=\frac{1-\cos\left(3+2\pi n\right)}{2\pi n+3}-\frac{1-\cos\left(3-2\pi n\right)}{2\pi n-3}\\ &=\frac{1-\cos\left(3\right)}{3+2\pi n}-\frac{1-\cos\left(3\right)}{3-2\pi n}\\ &=\left(1-\cos\left(3\right)\right)\left(\frac{1}{3+2\pi n}-\frac{1}{3-2\pi n}\right)\\ &=\left(1-\cos\left(3\right)\right)\left(\frac{3-2\pi n}{(3+2\pi n)(3-2\pi n)}-\frac{3+2\pi n}{(3-2\pi n)(3+2\pi n)}\right)\\ &=\left(1-\cos\left(3\right)\right)\left(\frac{-4\pi n}{9-4\pi^2n^2}\right)\\ &=\left(1-\cos\left(3\right)\right)\left(\frac{4\pi n}{4\pi^2n^2-9}\right)\color{red} \checkmark\\ \end{align*}\]

    • one year ago
  20. scarydoor
    Best Response
    You've already chosen the best response.
    Medals 2

    cool. That looks reasonable! Well done. I didn't help that much. I'm terrible at these trigonometric identities...

    • one year ago
  21. AccessDenied
    Best Response
    You've already chosen the best response.
    Medals 0

    It's great that you caught that little detail. It's always the little details... :P

    • one year ago
  22. UnkleRhaukus
    Best Response
    You've already chosen the best response.
    Medals 2

    if you wanted some context,

    • one year ago
    1 Attachment
    • Attachments:

See more questions >>>

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
  • 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.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.