1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Need help with this indefinite integral question please.

  1. Oct 9, 2016 #1
    1. The problem statement, all variables and given/known data
    find the following integral:

    cos(x/2) - sin(3x/2) dx


    2. Relevant equations
    I think the substitution method has to be used.
    Solve integrals by parts.

    3. The attempt at a solution
    Let u = x/2
    cosu

    du/dx=1/2, I then inverted it so dx/du = 2/1 = 2
    So dx=2du

    Now I have cosu2du

    Do I take out the constant ie: constant multiple rule?

    I am not confident about where to go next.

    Any help would be vastly appreciated, thank you.
     
    Last edited by a moderator: Oct 9, 2016
  2. jcsd
  3. Oct 9, 2016 #2

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Please turn off the bold font; it looks like you are yelling at us and is super distracting.
    Mod note: Fixed...I believe this is a matter of the formatting of the three numbered sections being inadventently carried over to the following text.
    Of course ##\int c f(u) \, du = c \int f(u) \,du## for any function ##f(u)## and any constant ##c##.
     
    Last edited by a moderator: Oct 9, 2016
  4. Oct 9, 2016 #3
    You should substitute not only in the cosine argument, but also in sine, so after the substitution you are left with [tex]\int 2(cos(u) - sin(3u))du[/tex] right?
    Now you are just a little step away from your result; integrals have linearity so yes, you can take the 2 out of the integral and easily calculate the primitives of the two terms.
     
  5. Oct 9, 2016 #4
    Sorry I was using the template and that was how it came out, it's my first post but I will make sure it is turned of from now on. Thank you very much for the help, helps to have a bit of confidence as I am doing this as an open uni course.
     
  6. Oct 9, 2016 #5
    Oh thank you very much mastrofoffi, I was reluctant to go on to the second part as I was concerned about the first but that makes a lot of sense.
     
  7. Oct 9, 2016 #6

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    The OP could also write ##\int [\cos(x/2)-\sin(3x/2)] \, dx = \int \cos(x/2) \, dx - \int \sin(3x/2) \, dx##, and use two separate substitutions: ##u = x/2## in the first integral and ##u = 3x/2## in the second one.
     
  8. Oct 9, 2016 #7

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Note that you can always check the answer to an indefinite integral by differentiating the answer to check you get the original function back.

    In principle, you should never be unsure whether you have the right answer.

    Using this approach you could also check for yourself the linearity of integrals and that, in particular, you can take a constant outside the integral.
     
  9. Oct 9, 2016 #8
    Okay I have had a good crack at it, my answer is:

    2sin(x/2) - 2/3(-cos 3x/2) + c

    Any thoughts?
     
  10. Oct 9, 2016 #9

    Mark44

    Staff: Mentor

    Did you read PeroK's post?
     
  11. Oct 9, 2016 #10

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    My thought is that you should differentiate that and see what you get.
     
  12. Oct 9, 2016 #11

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    At least you read it!
     
  13. Oct 9, 2016 #12
    Sorry I missed that post, so many posts, I feel this is going okay as my first ever forum though. I will now differentiate.
     
  14. Oct 9, 2016 #13

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Good. That is something you should get into the habit of doing automatically, always.
     
  15. Oct 9, 2016 #14
    Differentiated and got -sin(3x/2) so yeah looks good. Thank you very much for the help, I really appreciate it.
     
  16. Oct 9, 2016 #15
    Yeah makes complete sense as it's the reverse of integration.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Need help with this indefinite integral question please.
Loading...