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Thread: Integration

  1. #1 Integration 
    Forum Ph.D. Heinsbergrelatz's Avatar
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    can anyone help me integrate this function?



    i know you have to solve using substitution, but cant get around to it, any help i would sure;y appreciate. thank you in advance.


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  3. #2  
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    try partial fraction decomposition.


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  4. #3  
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    and it almost looks like a trigonometric substitution.
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  5. #4 Re: Integration 
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    Quote Originally Posted by Heinsbergrelatz
    can anyone help me integrate this function?



    i know you have to solve using substitution, but cant get around to it, any help i would sure;y appreciate. thank you in advance.
    Listen to salsaonline. His suggestion should work.
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  6. #5 Re: Integration 
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    Quote Originally Posted by Heinsbergrelatz
    can anyone help me integrate this function?



    i know you have to solve using substitution, but cant get around to it, any help i would sure;y appreciate. thank you in advance.
    Integration by partial fractions will work. Your evaluated integral should contain the product of a difference of logarithms and a constant.
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  7. #6  
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    ah yes thank you for all the help, but im still confused in whether to substitute by leaving or use partial fraction decomposition.
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  8. #7  
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    Quote Originally Posted by Heinsbergrelatz
    ah yes thank you for all the help, but im still confused in whether to substitute by leaving or use partial fraction decomposition.
    Then try them both.

    This is not that hard.
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  9. #8  
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    Quote Originally Posted by Heinsbergrelatz
    ah yes thank you for all the help, but im still confused in whether to substitute by leaving or use partial fraction decomposition.
    This is amongst the great difficulties of integration; it is often ambigious as to which method one should go about using. As Dr. Rocket said, simply try both methods to see which one will work; after you have done enough of these problems you will have a much greater sense of what technique to use; often, when I see an expression that's the derivative of another expression, U-substitution is a technique that I consider, for example.
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  10. #9  
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    This is not a case where you have two possible methods to choose from. You have one method, but it involves two steps. First: partial fraction decomposition. Second, u-substitution for the 1/(x^2 +16) part.
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  11. #10  
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    Quote Originally Posted by salsaonline
    This is not a case where you have two possible methods to choose from. You have one method, but it involves two steps. First: partial fraction decomposition. Second, u-substitution for the 1/(x^2 +16) part.
    But it is a case where if you think you have alternatives, you can try one and if it does not get you anywhere then you try something else. It beats indecision and inaction.

    Back when I was in school Eugene Wigner told a friend "One mark of genius is that when you try something and it does not work, the next time you try something else."

    Part of the learning process is trying things that don't work. Repeating the same mistakes is the sign of a slow learner.
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  12. #11  
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    Then try them both.

    This is not that hard.
    i tried both ways, and yes partial decomposition way was a cleaner or should i say a better arranged answer, than the one using the trigonometric substitution.
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  13. #12  
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    Quote Originally Posted by DrRocket
    Quote Originally Posted by salsaonline
    This is not a case where you have two possible methods to choose from. You have one method, but it involves two steps. First: partial fraction decomposition. Second, u-substitution for the 1/(x^2 +16) part.
    But it is a case where if you think you have alternatives, you can try one and if it does not get you anywhere then you try something else. It beats indecision and inaction.

    Back when I was in school Eugene Wigner told a friend "One mark of genius is that when you try something and it does not work, the next time you try something else."

    Part of the learning process is trying things that don't work. Repeating the same mistakes is the sign of a slow learner.
    I meant that you and I weren't suggesting two different approaches, but rather, one approach that involved two steps.
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  14. #13  
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    Quote Originally Posted by salsaonline
    Quote Originally Posted by DrRocket
    Quote Originally Posted by salsaonline
    This is not a case where you have two possible methods to choose from. You have one method, but it involves two steps. First: partial fraction decomposition. Second, u-substitution for the 1/(x^2 +16) part.
    But it is a case where if you think you have alternatives, you can try one and if it does not get you anywhere then you try something else. It beats indecision and inaction.

    Back when I was in school Eugene Wigner told a friend "One mark of genius is that when you try something and it does not work, the next time you try something else."

    Part of the learning process is trying things that don't work. Repeating the same mistakes is the sign of a slow learner.
    I meant that you and I weren't suggesting two different approaches, but rather, one approach that involved two steps.
    I agree. It did even occur to me that anyone would think of them as different approaches, particularly since my first post on the subject was basically "do what salsaonline said".

    My point was that Heinbergrelatz could learn by trying something that would not work, or that would be really messy (I don't see the trig substitution doing any good). That might him to understand why the method that you suggested does work, and work pretty cleanly at that.

    I think that one thing that does usually come through in elementary calculus classes is that if one simply picks up a pencil and writes down some function, the likelihood of being able to integrate it in closed form is not all that high, and if it is possible it may be pretty difficult to find the expression. That is why giant integral tables, like the one by Gradshteyn and Ryshik exist.

    If there is a problem taken from an elementary calculus book the likelihood is that it can be evaluated with one or two fairly simple techniques, or else it would not be in the book. If the integral comes up in the course of some non-contrived problem it may well be intractable.

    Once someone in the math department got a phone call from a contractor, wanting to know how to evaluate an integral, that it turned out was the area of a section of an ellipsoid. Needless to say one cannot evaluate that integral exactly in closed form. Now the natural question is why some construction contractor would want to evaluate such a thing. Ans. -- he had a contract to cover the New Orleans Superdome.
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    Quote Originally Posted by Heinsbergrelatz
    Then try them both.

    This is not that hard.
    i tried both ways, and yes partial decomposition way was a cleaner or should i say a better arranged answer, than the one using the trigonometric substitution.
    Could you show us what you got? I want to know if I've done it right
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  16. #15  
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    Could you show us what you got? I want to know if I've done it right
    you mean the answer to the solution, or the working?
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  17. #16  
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    Quote Originally Posted by Heinsbergrelatz
    Could you show us what you got? I want to know if I've done it right
    you mean the answer to the solution, or the working?
    The answer would be fine, though there's always more than one way to do these things...
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  18. #17  
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    ok here is the answer key from the book;

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  19. #18 Re: Integration 
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    Quote Originally Posted by Heinsbergrelatz
    can anyone help me integrate this function?



    i know you have to solve using substitution, but cant get around to it, any help i would sure;y appreciate. thank you in advance.



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  20. #19  
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    Dude, there's an edit button... and a preview button... and a delete button.
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  21. #20  
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    Quote Originally Posted by Heinsbergrelatz
    ok here is the answer key from the book;

    I get
    Wise men speak because they have something to say; Fools, because they have to say something.
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  22. #21  
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    Same thing. Multiply the constant by 2, and the log by 1/2, which is the same as taking the square root of what's in the log.
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  23. #22  
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    i have actually solved this question without using partial fractions method. See if it seems right;


    you firstly leave;



    =

    it simplifies down in to;

    =

    =

    now we must do a substitution one more time leaving-



    therefore the answer to the integral would be;



    now it gets familiar from here;



    now if we turn back on our first substitution, it was
    so..;



    so the final solution woould be;

    =
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  24. #23  
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    Quote Originally Posted by MagiMaster
    Same thing. Multiply the constant by 2, and the log by 1/2, which is the same as taking the square root of what's in the log.
    Oh yeah... It's been a while since I've dealt with logs, forgot about that property
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