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Integration Question

  1. Feb 19, 2008 #1
    I've been working on this problem for almost 4 days now and have made no progress. Once I think I've got it right, by professor says its wrong, and to try again. I've tried and tried. Any ideas? Here it is:

    [tex]\int sin^{3}x cos^{2}x dx[/tex]
  2. jcsd
  3. Feb 19, 2008 #2
    You could write everthing in the integral in terms of sines, then either:
    1. Use reduction formulae for sin^n x
    2. write the powers of sines in terms of sines of multiples of x
  4. Feb 19, 2008 #3
    Try to write the integrand as sin(x) * f(cos(x)) or cos(x) * f(sin(x)), where f is some algebraic function.
  5. Feb 19, 2008 #4
    I know this is not so formal, but it works:
    using the fact that:


    you get:

    [tex]\int sin^{3}(x) cos^{2}(x) dx=-\int sin^2(x)cos^2(x)d(cos(x))=-\int(1-cos^2(x))cos^2(x)d(cos(x))[/tex]

    and you are done.

  6. Feb 19, 2008 #5
    Thanks marco but I need the whole integral solved so the [tex]\int[/tex] sign is removed. And to the point where our constant C is added on: [tex]+ C[/tex].
  7. Feb 19, 2008 #6
    Do you honestly expect others to just do your work for you. Marco made the problem much simpler for you all it requires now is a simple, and fairly obvious substitution.
  8. Feb 19, 2008 #7
    No, the last part just doesn't make sense. If someone could explain it I'll take a shot at it, but I've never seen it (d(cos(x)))
    Last edited: Feb 19, 2008
  9. Feb 19, 2008 #8
    do the substitution:

    t=cos(x)-----> dt=d(cos(x))
    you get it???

  10. Feb 19, 2008 #9
    Thanks Marco.

    This is what I have so far, in continuation of what Marco helped out with:
    [tex]=-\int(\frac{1}{2}-\frac{1}{2} cos2x)(\frac{1}{2}+\frac{1}{2} cos2x)d(cos(x))[/tex]
    [tex]=-[{(\frac{1}{2}x-\frac{1}{4} sin2x)(\frac{1}{2}x+\frac{1}{4} sin2x)] + C[/tex]

    Err, is this right? If not how do I fix it? Thanks
  11. Feb 19, 2008 #10
    No, that's not right.

    Do what macro_84 said. let t = cos(x)

    Now substitute t everywhere you see a cos(x) in the integrand that macro gave you (the one with nothing but cosines). it's staring you in the face.
  12. Feb 19, 2008 #11
    Err... That did not make much sense, sorry.
  13. Feb 19, 2008 #12
    How did that not make sense? Are you familiar with integration by substitution? Make the substititution t=cos(x) and what happens?
  14. Feb 19, 2008 #13
    tx = -sin(x)?
  15. Feb 19, 2008 #14


    User Avatar

    Well no wonder you couldn't get this integral, you don't understand the most basic method of integration.
  16. Feb 20, 2008 #15
    Sorry, I've only been doing this for a few weeks. I came here for help, nothing else.
  17. Feb 20, 2008 #16
  18. Feb 20, 2008 #17

    can you do it now??

  19. Feb 20, 2008 #18


    User Avatar
    Staff Emeritus
    Science Advisor

    Then stop expecting people to do the problem for you. What has been suggested is that you rewrite the integral as
    [tex]\int sin^2(x)cos^2(x) (sin(x)) dx= \int (1- cos^2(x))cos^2(x) (sin(x)dx[/itex]
    Now, if u= cos(x), what is du? If you don't know that you should review differentiation before trying integration.
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