A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

anonymous

  • 5 years ago

how do i solve the integral of sin(ax) cos(ax)? i don't understand the method of goniometric integrals

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

    First, let u=ax and du=a dx. dx = du/a, so you can simplify it to being 1/a * ∫ sin(u)cos(u) du. Now, I'm basically going to do another u-sub, but for clarity going to call it v: v = cos(u), dv = -sin(u) du. Now, du = -dv/sin(u), and when you plug the very last phrase into your integral, you have -1/a * ∫v dv. Then, solve and re-substitute: -1/a * v^2/2 + c = -1/a * cos(u)^2/2 + c = -1/(2a) * cos(ax)^2 + c. Not really sure what you mean by "goniometric" though...

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

    why do you do du=a dx?

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

    Just to get the 1/a in front.

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

    I don't think it's absolutely necessary, but it makes things a little cleaner during your second substitution. The absolutely crucial sub here is v = cos(u), dv = -sin(u) du because that eliminates sin(u) from your integral.

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

    nice, you made me get it, I hope i can put this to practice, thanks!

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

    No problem; I'm glad I could help. :)

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

    what i meant with goniometric, my study method does something else: F sin ax cos ax dx = 1/a F cos ax d(-cos ax) = -1/a F cos ax d cos ax = -1/2a cos^2 ax +C i understand your method, just not the one we are supposed to know

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

    an easy method is u=ax then:1/a * ∫ sin(u)cos(u) du so sin(u)cos(u)=(1/2)*sin(2u) and then the integral equals (1/2a )* ∫ sin(2u)du=(-1/4a)*cos(2u)+c .

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

    ijadi, I'm not sure why but according to Wolfram|Alpha and another computerized integrator our answers are not congruent, even though I can't see anything wrong with your method. Strange. Meester, the only thing I meant was that I've never heard of goniometric before in my life! xD I don't see exactly why that name is attached but I understand how you did that integral now.

  10. 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.