A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

anonymous

  • 5 years ago

integrate from 0 to pi/2 (sinx + cosx)/(9+16sin2x) dx

  • This Question is Closed
  1. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Recall sin 2x= 2[[(cos)^2]x]-1

  2. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yup.. what nxt?

  3. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Choose your u.

  4. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    can you solve it out buddy.. i seem to get nowhere!

  5. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    You have to make an attempt. Don't get overwhelm. Your task is simple: choose a u. You could be right, you could be wrong. You won't harm anything.

  6. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[\sin x+\cos x \over 9+16(2\cos ^{2}x-1)\] \[\sin x+ \cos x \over 32\cos ^{2}x-7\] what do you think i should select as u

  7. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    We're not getting anywhere. I asked you twice: choose a u.

  8. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i have no idea buddy.. i ll try later

  9. watchmath
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I am curious how we can really go from there. BTW \(\sin(2x)\neq 2\cos^2(x)-1\). But \(\cos(2x)=2\cos^2(x)-1\)

  10. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    lol... i didnt evn realize that! rats!

  11. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Good catch, watchmath. Apparently this question appears frequently in India exams. Solution depends on the relation 1 - sin 2x=(sinx - cos x)^2. Let sin x - cos x = t (cos x + sin x) dx = dt t^2= (sin x - cos x)^2 = 1 - sin 2x sin 2x = 1-t^2.\[\int\limits_{0}^{\pi/2}(\sin x + \cos x)/(9+16 \sin 2x)\]=\[\int\limits_{0}^{\pi/2}dt/(9+16(1-t ^{2})\]

  12. watchmath
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    http://openstudy.com/users/watchmath#/users/watchmath/updates/4dd3bbd699508b0bc2d6a5a4

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

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

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.