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Integral Help!

  1. Aug 8, 2011 #1
    ∫1+sinx/(cosx)² dx

    I made u = cos x.

    du = -sinxdx

    -du = sinxdx

    so:

    ∫1+sinxdx/(cosx)² = ∫1-du/u² = ∫ (1/u²) * (1-du)

    This is where I got stuck. the 1-du is throwing me off. distributing would get me nowhere and I don't know how to get rid of the 1.

    Please help!
     
  2. jcsd
  3. Aug 8, 2011 #2
    We have a https://www.physicsforums.com/forumdisplay.php?f=152"!

    Is the integrand (1 + sin x)/cos2x ? Or 1 + sin x/cos2x?

    In the second case, for the second addend note that if u = cos x, -du = sin x dx:

    ∫ 1 + sin x/cos2x dx = x + ∫ -du/u2 = x + 1/u + C = x + 1/cos x + C

    In the first case, write 1/cos2x + sin x/cos2x and integrate the second term as we did previously.
    Also, 1/cos2x = sec2x -- this is the derivative of which function?
     
    Last edited by a moderator: Apr 26, 2017
  4. Aug 8, 2011 #3
    first case, and ill remember that next time, although this isn't homework
     
  5. Aug 8, 2011 #4
    Once I get it to ∫1-du/u²

    If I separate it:

    [∫(1/u²)]-[∫(du/u²)]

    [∫(1/u²)du]-[∫(1/u²)du]

    This = 0

    I can only separate it at the beginning?
     
  6. Aug 8, 2011 #5

    rock.freak667

    User Avatar
    Homework Helper

    Instead of substitution where you would get something like dx = 1-du, which is not helpful, why don't we rewrite the integral as

    [tex]\int \left( \frac{1}{cos^2x} + \frac{sinx}{cos^2x} \right) dx[/tex]

    Now do you know what 1/cosx is the same as and what sinx/cosx is the same as?

    When you get that, rewrite it the integral again but replace 1/cosx with the equivalent and do the same with sinx/cosx.

    Post what you get.
     
  7. Aug 8, 2011 #6

    Mark44

    Staff: Mentor

    This is incorrect, in part because it is meaningless.
    Each integral needs the differential factor. In what you have above, ∫(1/u²) is missing du.
     
  8. Aug 8, 2011 #7
    ah ok, now I understand.

    Thank you for all the help guys! :D
     
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