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Math Help - Exam question - Integrals

  1. #1
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    Exam question - Integrals

    The results of my calculus exam came and there is a question I had wrong.

    Which is the integral of (sin(logx))/x?

    My answer was -cos(logx) + c

    I checked wolfram and the answer is the same, therefore that makes my answer correct right?
    However, it says on the paper that this answer is wrong.
    I did some research and realised that the derivative of -cos(logx) is not (sin(logx))/x, and maybe that's why it is wrong.
    But how come I did not get the correct answer? Even wolfram says my answer is correct.

    My resoltuion process was the following:

    - turn (logx) into u;
    - turn du into (1/x)dx -------> dx = xdu

    - that gives me the integral of (xsen(u)/x);
    - cutting the x's on either side gives me the integral of sen(u)
    - resolving the integral gives me -cos(u)
    - replacing the u gives me -cos(logx) and the add the constant +c

    This method always never failed me, so what went wrong?

    I have to go to college tomorrow to defend my answer, so I would like to get some replies fast... I need two points more on this exam to pass this subject.

    Thank you.
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  2. #2
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    the derivative of -cos(logx) is not (sin(logx))/x
    It's not?
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  3. #3
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    It gives me sin(logx);
    The original function multiplies that for 1/x
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  4. #4
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    You're forgetting the chain rule.
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  5. #5
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    I agree with both of you.

    @CMartins: What answer did your teacher give you?

    -Dan
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    @topsquark: I didn't talk to him yet. He just posted on the college website the results of each question. And apparently, I got that one wrong. I have to go there tomorrow to review my exam, that is why I need to prepare to defend my answer.

    @Ackbeet: You're right! In that case my answer is correct right?
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  7. #7
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    I would say your answer is correct. You should have been looking to use the chain rule to verify the integration, if the rule of integration used was a u substitution (they are inverses, just like by-parts is the inverse of the product rule).
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  8. #8
    Forum Admin topsquark's Avatar
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    I can't get anywhere with the idea but could this be a case of the arbitrary constant is different between the exam answer and CMartins's answer?

    -Dan
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  9. #9
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    Quote Originally Posted by topsquark View Post
    I can't get anywhere with the idea but could this be a case of the arbitrary constant is different between the exam answer and CMartins's answer?

    -Dan
    I don't know. It seems a fairly straight-forward problem to me. My guess is that the professor's key is wrong. There's a typo there or something.

    By the way, I enjoyed the Dune quote in your sig.
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  10. #10
    Forum Admin topsquark's Avatar
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    It's just about the only thing from Dune that can actually be applied in the real world. It's a nice little piece for a simple meditative state when you are nervous.

    -Dan
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  11. #11
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Ackbeet View Post
    I don't know. It seems a fairly straight-forward problem to me. My guess is that the professor's key is wrong. There's a typo there or something.
    It probably is simply a typo on the answer key, but I have a vague memory of a problem involving logs where two answers were essentially the same, but the arbitrary constant made the two solutions look like they were different. But I don't remember the details.

    -Dan
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  12. #12
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    Thank you. I will talk to him tomorrow and resolve this.
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  13. #13
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    You're very welcome for my contribution. Have fun!
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    Quote Originally Posted by topsquark View Post
    It probably is simply a typo on the answer key, but I have a vague memory of a problem involving logs where two answers were essentially the same, but the arbitrary constant made the two solutions look like they were different. But I don't remember the details.
    Was it a case of the following? We have that I = \int\frac{1}{ax+b}\;{dx} = \frac{1}{a}\ln\left|ax+b\right|+k_{1}. But also have
    that I = \int\frac{1}{ax+b}\;{dx} = \frac{1}{a}\int\frac{1}{x+\frac{b}{a}}\;{dx} = \frac{1}{a}\ln\left|x+\frac{b}{a}\right|+k_{2}. These are really the same because:
    \frac{1}{a}\ln\left(ax+b\right)+k_{1}  = \frac{1}{a}\ln\left|a\left(x+\frac{b}{a}\right)\ri  ght|+k_{1}  =\frac{1}{a}\ln\left|x+\frac{b}{a}\right|+\frac{1}  {a}\ln{a}+k_{1} = \frac{1}{a}\ln\left|x+\frac{b}{a}\right|+k_{2}.
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  15. #15
    Forum Admin topsquark's Avatar
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    Yes, that might very well be the one I was thinking of. Thanks!

    -Dan
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