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I really don't get the substitution rule

  1. Feb 1, 2007 #1


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    I really don't get the substitution rule. This is supposed to be the easiest problem in the homework set: u=3x
    [tex]\int {\cos \,3x\,\,dx\,\, = \,\,\int {\cos \,u\,\, = \,\,\sin \,u + C\,\, = \,\,\sin 3x + C} } [/tex]

    But the right answer is 1/3 sin(3x). Where did the 1/3 come from?
  2. jcsd
  3. Feb 1, 2007 #2
    Look at your second integral in here:
    [tex]\int {\cos \,3x\,\,dx\,\, = \,\,\int {\cos \,u\,\,dx???\,\, = \,\,\sin \,u + C\,\, = \,\,\sin 3x + C} } [/tex]

    with respect to what variable? x?
    or, did you mean to have a du in there?

    If you meant for a du to be there, what makes you think that du can be substituted for dx? You wrote u=3x. if I take the derivative, with respect to x, I get du/dx = 3. Multiplying both sides by dx (thanks, Leibnitz for that notation), I get du = 3 dx. And from that, dx will be equal to 1/3 du
  4. Feb 1, 2007 #3
    Actually, I think a "harder" integral will make this easier to see:
    [tex]\int {\sec x \sec x \tan xdx [/tex]
    Let u = secx
    du/dx = secxtanx
    du = secxtanxdx

    Now, when you substitute, u will take care of the first secx.
    But, du will take care of the rest of it: secxtanxdx
  5. Feb 1, 2007 #4


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    I think I did it right, but I still have a question.

    [tex]u = 3x,\,\,\,\,\frac{{du}}{{dx}} = \left( {3x} \right)^\prime = 3,\,\,\,\,dx = \frac{{du}}{\begin{array}{l}
    3 \\
    \end{array}},\,\,\,\,\int {\cos \,3x\,\,dx\,\,} = \,\,\int {\cos \,u\,\,\frac{{du}}{3} = \frac{{\sin u}}{3}} = \frac{{\sin 3x}}{3}[/tex]

    So when I integrate, the du simply disappears?

    I hate Leibnitz notation. I do not think it is intuitive. I wish in Calc 1 they would have spent a lecture on Leibnitz notation. Rather, they just started using it without describing it.
    Last edited: Feb 1, 2007
  6. Feb 1, 2007 #5
    Yes, the du disappears with the integral sign (as would dx, dy, etc). Kind of like: 4 + 3 = 7, we don't leave the + there.
  7. Feb 1, 2007 #6
    [tex]\int {\cos \,u\,\,\frac{{du}}{3}[/tex]

    At this point you should factor out the constant, 1/3, from the integral:

    [tex]\frac{1}{3} \int {\cos \,u\,\,du[/tex]
  8. Feb 2, 2007 #7


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    :rofl: Thanks to you, I will never forget that.

    Thanks, everyone. :rolleyes: This was so confusing to me until I asked here. I just plowed through the next 10 problems with ease. But I'm stuck on the 11th, so I'm giving up for the night.
  9. Feb 2, 2007 #8
    Man, that was an easy problem. Think forward of the answer. I guess light years will come up for your mind to catch up.
  10. Feb 2, 2007 #9


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    It's only easy if you know the answer! When learning, any problem can be difficult if you can't see what to do!
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