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Math Help - The addition-subtraction puzzle (calculus)

  1. #1
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    The addition-subtraction puzzle (calculus)

    In my easy calculus puzzle, I've submitted \dfrac{1}{\sin(x)cos(x)} (in another form) and I pointed out how by letting 1 = sin^2x + cos^2x would lead to a solution. The formula can be rewritten as 1 = (1 -cox^2x) + cos^2x for the numerator of the first fraction which of course leads to the same answer.

    That method I refer to as the addition-subtraction method or ASM for short (which is applied to the numerators of fractions). This is a method that should be taught in calculus classes and demonstrated in texts, but unfortunately isn't. It's a powerful method for simplifying fractions and I'll now give you another example in the form of an easy puzzle.

    Using ASM, transform \dfrac{1}{x(1 + x^2)} into two fractions and then go on to integrate those two fractions.
    Last edited by wonderboy1953; January 12th 2011 at 12:00 PM. Reason: clarification
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  2. #2
    Member Pranas's Avatar
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    Quote Originally Posted by wonderboy1953 View Post
    In my easy calculus puzzle, I've submitted \dfrac{1}{\sin(x)cos(x)} (in another form) and I pointed out how by letting 1 = sin^2x + cos^2x would lead to a solution. The formula can be rewritten as 1 = (1 -cox^2x) + cos^2x for the numerator of the first fraction which of course leads to the same answer.

    That method I refer to as the addition-subtraction method or ASM for short (which is applied to the numerators of fractions). This is a method that should be taught in calculus classes and demonstrated in texts, but unfortunately isn't. It's a powerful method for simplifying fractions and I'll now give you another example in the form of an easy puzzle.

    Using ASM, transform \dfrac{1}{x(1 + x^2)} into two fractions and then go on to integrate those two fractions.
    I believe your idea is
    \displaystyle \[\frac{1}{{x(1 + {x^2})}} = \frac{{(1 + {x^2}) - {x^2}}}{{x(1 + {x^2})}} = \frac{{1 + {x^2}}}{{x(1 + {x^2})}} - \frac{{{x^2}}}{{x(1 + {x^2})}} = \frac{1}{x} - \frac{x}{{1 + {x^2}}}\]

    Hence
    \displaystyle \[\int {\frac{{dx}}{{x(1 + {x^2})}}}  = \int {(\frac{1}{x} - \frac{x}{{1 + {x^2}}}) \cdot dx}  = \int {\frac{{dx}}{x}}  - \int {\frac{{x \cdot dx}}{{1 + {x^2}}}}  = \]
    \displaystyle \[ = \ln \left| x \right| - \frac{1}{2} \cdot \ln \left| {{x^2} + 1} \right| + C = \ln \left| {\frac{x}{{\sqrt {{x^2} + 1} }}} \right| + C\]
    Last edited by Pranas; January 14th 2011 at 06:52 AM.
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  3. #3
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    \displaystyle \int{\frac{1}{\sin{x}\cos{x}}\,dx} = \int{\frac{\cos{x}}{\sin{x}\cos^2{x}}\,dx}

    \displaystyle = \int{\frac{\cos{x}}{\sin{x}(1 - \sin^2{x})}\,dx}

    \displaystyle = \int{\frac{\cos{x}}{\sin{x}(1 - \sin{x})(1 + \sin{x})}\,dx}.


    Now make the substitution \displaystyle u = \sin{x} so that \displaystyle \frac{du}{dx} = \cos{x} and the integral becomes

    \displaystyle \int{\frac{1}{u(1 - u)(1 + u)}\,\frac{du}{dx}\,dx}

    \displaystyle = \int{\frac{1}{u(1 - u)(1 + u)}\,du}

    \displaystyle = \int{\frac{1}{u} - \frac{1}{2(u - 1)} - \frac{1}{2(u + 1)}\,du}

    \displaystyle = \ln{|u|} - \frac{1}{2}\ln{|u-1|} - \frac{1}{2}\ln{|u + 1|} + C

    \displaystyle = \ln{|\sin{x}|} - \frac{1}{2}\ln{|\sin{x} - 1|} - \frac{1}{2}\ln{|\sin{x} + 1|} + C.
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  4. #4
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    Quote Originally Posted by Prove It View Post
    \displaystyle \int{\frac{1}{\sin{x}\cos{x}}\,dx} = \int{\frac{\cos{x}}{\sin{x}\cos^2{x}}\,dx}

    \displaystyle = \int{\frac{\cos{x}}{\sin{x}(1 - \sin^2{x})}\,dx}

    \displaystyle = \int{\frac{\cos{x}}{\sin{x}(1 - \sin{x})(1 + \sin{x})}\,dx}.


    Now make the substitution \displaystyle u = \sin{x} so that \displaystyle \frac{du}{dx} = \cos{x} and the integral becomes

    \displaystyle \int{\frac{1}{u(1 - u)(1 + u)}\,\frac{du}{dx}\,dx}

    \displaystyle = \int{\frac{1}{u(1 - u)(1 + u)}\,du}

    \displaystyle = \int{\frac{1}{u} - \frac{1}{2(u - 1)} - \frac{1}{2(u + 1)}\,du}

    \displaystyle = \ln{|u|} - \frac{1}{2}\ln{|u-1|} - \frac{1}{2}\ln{|u + 1|} + C

    \displaystyle = \ln{|\sin{x}|} - \frac{1}{2}\ln{|\sin{x} - 1|} - \frac{1}{2}\ln{|\sin{x} + 1|} + C.
    How about

    \displaystyle \[\int {\frac{1}{{\sin x\cos x}}{\mkern 1mu} dx}  = \int {\frac{{\cos x}}{{\sin x{{\cos }^2}x}}{\mkern 1mu} dx}  = \int {\frac{{d(\tan x)}}{{\tan x}}{\mkern 1mu} }  = \ln \left| {\tan x} \right| + C\]

    Since
    \displaystyle \[d(\tan x) = \frac{{dx}}{{{{\cos }^2}x}}\]
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  5. #5
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    Very good

    Quote Originally Posted by Pranas View Post
    I believe your idea is
    \displaystyle \[\frac{1}{{x(1 + {x^2})}} = \frac{{(1 + {x^2}) - {x^2}}}{{x(1 + {x^2})}} = \frac{{1 + {x^2}}}{{x(1 + {x^2})}} - \frac{{{x^2}}}{{x(1 + {x^2})}} = \frac{1}{x} - \frac{x}{{1 + {x^2}}}\]

    Hence
    \displaystyle \[\int {\frac{{dx}}{{x(1 + {x^2})}}}  = \int {(\frac{1}{x} - \frac{x}{{1 + {x^2}}}) \cdot dx}  = \int {\frac{{dx}}{x}}  - \int {\frac{{x \cdot dx}}{{1 + {x^2}}}}  = \]
    \displaystyle \[ = \ln \left| x \right| - \frac{1}{2} \cdot \ln \left| {{x^2} + 1} \right| + C = \ln \left| {\frac{x}{{\sqrt {{x^2} + 1} }}} \right| + C\]
    Exactly what I had in mind. Normally this problem would be solved by the method of partial fractions which involves some algebra, but this way is much more efficient (wouldn't you agree?)
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  6. #6
    Member Pranas's Avatar
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    Quote Originally Posted by wonderboy1953 View Post
    Exactly what I had in mind. Normally this problem would be solved by the method of partial fractions which involves some algebra, but this way is much more efficient (wouldn't you agree?)
    Indeed, foreseeing the possibilities is always a good thing in mathematics
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