1. Limited time only! Sign up for a free 30min personal 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!

Homework Help: Trig integration (cos(x)+sin(x)) / sin(2x)

  1. Oct 3, 2009 #1
    Prepping for my test, and I can't seem to solve this problem.

    1. The problem statement, all variables and given/known data

    [tex]\int \frac{cos(x) + sin(x)}{sin(2x)}dx[/tex]

    2. Relevant equations

    Not sure if it led me astray, but I used the trig. property:

    [tex]sin(2x) = 2sin(x)cos(x)[/tex]

    3. The attempt at a solution

    [tex]= \int \frac{cos(x) + sin(x)}{2sin(x)cos(x)}dx[/tex]

    [tex]= 1/2 \int(csc(x) + sec(x))dx[/tex]

    [tex]= 1/2ln|csc(x) - cot(x)| + 1/2ln|sec(x) + tan(x)| + C[/tex]

    [tex]= 1/2 ln|csc(x)sec(x) + sec(x) - csc(x) - 1| + C[/tex]

    I feel like this is wrong, but not sure where I'm messing up. Maybe my trig. property isn't the way to go.
  2. jcsd
  3. Oct 3, 2009 #2


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Looks good to me except you need + C.:cool:
  4. Oct 3, 2009 #3
    I would keep the answer in the form of the line before the last one. There's one little mistake in that line, though.
    ∫cscx dx = -ln|cscx + cotx| + C, not ln|cscx - cotx|
  5. Oct 3, 2009 #4


    User Avatar
    Homework Helper







    Now, lets assume:[tex]ln\left(\frac{sinx}{1+cosx}\right)=ln\left(\frac{1-cosx}{sinx}\right)[/tex]






    Therefore, they are equivalent.
  6. Oct 3, 2009 #5
    Graphically, it looks like they are the same if you ignore the absolute values. LHS - RHS looks like 0 on wolframalpha, but with absolute values, the graph of their difference is not 0.
  7. Oct 4, 2009 #6


    User Avatar
    Homework Helper

    I re-checked my work and didn't seem to find any errors or possibilities of extraneous solutions etc. so I graphed it as well, but on graphmatica. It seems that with or without absolute values, both functions are equal. If you believe you haven't made a mistake, please post a picture of what you see on wolfram :smile:
  8. Oct 4, 2009 #7
  9. Oct 4, 2009 #8
    Ok, there is no difference between our integrals of cscx. I didn't recognize the other form immediately (I never worked with ∫cscx dx by itself either) so I thought it was wrong. Looking at the table of integrals in my calc book, I see that same integral, ln|cscx - cotx| and I did find it by looking on the internet, but apparently that form is not as common as the other one.

    I learn something new almost every day on this site. :smile:
  10. Oct 4, 2009 #9
    cool discussion guys. I just worked the integral of csc(x) on my own, and I came up with that funky version. Didn't realize it was less common :).
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook