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Integrating trig

  1. Mar 7, 2006 #1
    integrate with respect to x: (sinx)^3 * cosx

    i have no idea where to start, can anyone help me? i've looked at differentials of other trig functions but i can't see any that would help :mad:
     
  2. jcsd
  3. Mar 7, 2006 #2

    benorin

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    Use a "u substitution".
     
  4. Mar 7, 2006 #3
    what do you mean?
     
  5. Mar 7, 2006 #4
    you want to use a u substitution. Find a value in your expression to be u, and find another one to be du.

    ~Lyuokdea
     
  6. Mar 7, 2006 #5

    benorin

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    How would you integrate [tex]\int (x^2+7)^3 \cdot 2x \,dx[/tex] ?
     
  7. Mar 7, 2006 #6
    Use U sub. as indicated before. Let your U = sin(x) . Work it from there
     
  8. Mar 8, 2006 #7

    VietDao29

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    If your sine function is raised to an odd power, it's commonly to let u = cos x, and work from there.
    If your cosine function is raised to an odd power, it's commonly to let u = sin x, and work from there.
    If both are raised to an odd power, then you can either let u = sin x, or u = cos x.
    Note that, you should sometimes need to use the Pythagorean identity: sin2x + cos2x = 1, to solve your problem.
    I'll give you an example:
    -----------
    Example:
    [tex]\int \cos x \sin ^ 2 x dx[/tex]
    cos x is raised to the power 1, hence it's an odd power, let u = sin x.
    u = sin x => du = cos x dx, right? Substitute that into your integral, we have:
    [tex]\int u ^ 2 du = \frac{u ^ 3}{3} + C[/tex]
    Change u back to x, gives:
    [tex]\int \cos x \sin ^ 2 x dx = \frac{\sin ^ 3 x}{3} + C[/tex]
    Can you go from here? :)
     
  9. Mar 8, 2006 #8
    i've never done anything like that before and i don't quite understand it
     
  10. Mar 8, 2006 #9
    Let me see if I can help you. :smile:

    ~Kitty
     
  11. Mar 8, 2006 #10
    Let me see if I have this correct, you have:

    sine of x cubed times cosine of x right?

    ~Kitty
     
  12. Mar 8, 2006 #11
    Please don't take this as attacking. Have you been exposed to u substitution? I'm assuming you have.

    ~Kitty
     
  13. Mar 8, 2006 #12
    My process was about the same as VietDao. I'm sorry.

    ~Kitty
     
  14. Mar 8, 2006 #13
    Too many posts... :\
    Dude, if you didn't take this in class, read off this site for help. :}
     
  15. Mar 11, 2006 #14
    unfortunately not. but it turns out i didn't have to do it after all - it was off-the-syllabus stuff
     
  16. Mar 11, 2006 #15
    As Viet told take u = cosx.
    Then find du/dxand the relation between both to replace du in place of dx in the integration
     
  17. Mar 12, 2006 #16

    arildno

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    Hmm..use of u=sinx is simpler in this case.
     
  18. Mar 15, 2006 #17
    the only u substitution i've used is in differentiation using the chain rule
     
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