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Integrate cosx^3dx

  1. Jan 24, 2004 #1
    Any suggestions on how to integrate
    [tex]\int cos^3xdx[/tex]
    My first clue would be to substitute
    (1-(Sinx)^2)cosxdx but any clue to go after this?
    Last edited: Jan 24, 2004
  2. jcsd
  3. Jan 24, 2004 #2


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    Well, do any of the techniques you've learned apply to

    [tex]\int (1 - \sin^2 x) \cos x \, dx[/tex]

  4. Jan 24, 2004 #3
    Put sinx=t => cosx dx=dt

    so u have int(1-t2)dt=t-t3/3
  5. Jan 24, 2004 #4
    thanks for the help. This trig integration stuff is killer

    OK the reason I asked was because I have a lab assignment where I have to calculate the powers of [tex]\int cos^n[x]dx[/tex] where n is odd numbers from 3-13.
    So can someone please look over my work for [tex]\int cos^5[x]dx[/tex] and make sure it looks valid:
    First I substituted [tex](1-sin^2[x])^2[/tex] for [tex]cos^4[x][/tex] which gives me:
    [tex]\int (1-sin^2[x])^2 cos[x]dx[/tex]
    Then I let u=sinx meaning du=cosxdx
    Giving me:
    [tex]\int (1-u^2)^2du[/tex]
    Which I simplfied to give me:
    [tex]\int u^4 du - \int 2u^2 du + \int du[/tex]
    Which gives me
    [tex](u^5)/5-(2u^3)/3 + u[/tex]

    Anyone see any glaring errors?
    Thank you in advance.
    Last edited: Jan 24, 2004
  6. Jan 24, 2004 #5


    is (1-u^2) du.
  7. Jan 25, 2004 #6


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    Looks right to me. (except you should substitute back what u is)
  8. Jan 25, 2004 #7
    How it is wrong

    Moreover General way will be to form a reduction formula
    Last edited: Jan 25, 2004
  9. Jan 25, 2004 #8
    Yeah I didn't feel like typing out any more latex:smile:
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