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Math Help - how to show that....

  1. #1
    ADY
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    how to show that....

    1) Using:

    \int (f(x))^n f'(x) dx = \frac{1}{n + 1}(f(x))^{n+1} + c

    how can i show...

    \int \frac{sin x}{(2 - cos x)^3} dx = - \frac{1}{2(2 - cos x)^2} + c

    where c is an arbitrary constant.

    2) Now in implicit form, find the solution

    \frac{dy}{dx} = \frac{6y^{2/3}sin x}{(2- cos x)^3}

    3) what is the corresponding answer, implicit that satisifies y = 1 when x = 0

    4) Whats the explicit form of the answer?

    5) whats y's value, given by the solution when  x = 1/2 \pi

    Thanks
    Last edited by ADY; January 8th 2010 at 09:20 AM.
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  2. #2
    Senior Member Shanks's Avatar
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    \int \frac{sin x}{(2 - cos x)^3} dx = \int \frac{1}{(2 - cos x)^3} d(2-cos x) = - \frac{1}{2(2 - cos x)^2} + c

    \frac{dy}{dx} = \frac{6y^{2/3}sin x}{(2- cos x)^3}
    implies
    \frac{dy}{6y^{2/3}}=\frac{sin x dx}{(2- cos x)^3}
    Integrate both sides to get the explicit form.


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  4. #4
    ADY
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    That helps with 1), any help with the others please?
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  5. #5
    MHF Contributor Amer's Avatar
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    2) \frac{dy}{dx} = \frac{6y^{\frac{2}{3}} \sin x}{(2-\cos x)^3 }

    \frac{dy}{6y^{\frac{2}{3}}} = \frac{\sin x\cdot dx}{(2-\cos x)^3}

    \int \frac{dy}{6y^{\frac{2}{3}}} = \int \frac{\sin x\cdot dx}{(2-\cos x)^3}

    the right side we find it in the first question

     \frac{y^{\frac{-2}{3} +1}}{6\left(\frac{-2}{3}+1\right)}=- \frac{1}{2(2 - cos x)^2} + c<br />

    \frac{y^{\frac{1}{3}}}{2} = - \frac{1}{2(2 - cos x)^2} + c<br />

    3)when y=1 and x=0

    \frac{1}{2} = \frac{-1}{2(2-1)^2}+c

    \frac{1}{2} = \frac{-1}{2} + c \Rightarrow c = 1 so

    \frac{y^{\frac{1}{3}}}{2} = \frac{-1}{2(2-\cos x)^2} + 1

    y^{\frac{1}{3}} = \frac{-1}{(2-\cos x)^2} + 2

    4)

    y = \left( \frac{-1}{(2-\cos x)^2} +2 \right)^3


    5)when x= \frac{1}{2\pi}

     y = \left( \frac{-1}{(2-\cos (1/2\pi))^2} +2 \right)^3
    Last edited by Amer; January 8th 2010 at 01:23 PM.
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  6. #6
    ADY
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    I thought i understood the first bit properly, but i don't.

    So we have:

    \int (f(x))^n f'(x) dx = \frac{1}{n + 1}(f(x))^{n+1} + c

    and i need to show that:

    \int \frac{sin x}{(2 - cos x)^3} dx = - \frac{1}{2(2 - cos x)^2} + c

    Where c is an arbitray constant.

    ____________________________

    I Answered it to show that:

    \int f^n f ' = \frac{1}{n + 1} f^{n+1}

    \int (2 - cos x)^{-3} sin x dx

    \frac{1}{-3+1} (2 . cos x)^{-3+1} + c

    Is this correct?
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  7. #7
    Senior Member Shanks's Avatar
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    a little error, pay attention to the operation between 2 and cos x.
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  8. #8
    ADY
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    ok but what am i missing because i feel the question doesnt finish there.

    after


    \frac{1}{-3+1} (2 . cos x)^{-3+1} + c


    what do i need to do next, this is important as it effects all the subsquent questions?
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  9. #9
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    The substitution rule is easier if written as

    \int{f\left(u(x)\right)\,\frac{du}{dx}\,dx} = \int{f(u)\,du}.


    So in your case

    \int{\frac{\sin{x}}{(2 - \cos{x})^3}\,dx} = \int{(2 - \cos{x})^{-3}\sin{x}\,dx}

    Let u = 2 - \cos{x} so that \frac{du}{dx} = \sin{x}.

    The integral becomes

    \int{u^{-3}\,\frac{du}{dx}\,dx} = \int{u^{-3}\,du}

     = -\frac{1}{2}u^{-2} + C

     = -\frac{1}{2}(2 - \cos{x})^{-2} + C

     = -\frac{1}{2(2 - \cos{x})^2} + C.
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  10. #10
    ADY
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    Ok great,

    so now with your solution has that changed the subsquent questions...

    2) Now in implicit form, find the solution

    \frac{dy}{dx} = \frac{6y^{2/3}sin x}{(2- cos x)^3}
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  11. #11
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    Quote Originally Posted by ADY View Post
    Ok great,

    so now with your solution has that changed the subsquent questions...

    2) Now in implicit form, find the solution

    \frac{dy}{dx} = \frac{6y^{2/3}sin x}{(2- cos x)^3}
    Well

    \frac{dy}{dx} = \frac{6y^{\frac{2}{3}}\sin{x}}{(2 - \cos{x})^3}

    y^{-\frac{2}{3}}\,\frac{dy}{dx} = \frac{6\sin{x}}{(2 - \cos{x})^3}

    Now integrate both sides with respect to x...

    \int{y^{-\frac{2}{3}}\,\frac{dy}{dx}\,dx} = \int{\frac{6\sin{x}}{(2 - \cos{x})^3}\,dx}

    \int{y^{-\frac{2}{3}}\,dy} = 6\int{\frac{\sin{x}}{(2 - \cos{x})^3}\,dx}.

    Can you go from here? Hint: the RHS is the same as Q1...
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  12. #12
    Senior Member Shanks's Avatar
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    Amer provide you the correct solution, you can refer it.
    If you have any question, don't hesitate to ask for help.
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  13. #13
    ADY
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    Quote Originally Posted by Prove It View Post
    Well

    \frac{dy}{dx} = \frac{6y^{\frac{2}{3}}\sin{x}}{(2 - \cos{x})^3}

    y^{-\frac{2}{3}}\,\frac{dy}{dx} = \frac{6\sin{x}}{(2 - \cos{x})^3}

    Now integrate both sides with respect to x...

    \int{y^{-\frac{2}{3}}\,\frac{dy}{dx}\,dx} = \int{\frac{6\sin{x}}{(2 - \cos{x})^3}\,dx}

    \int{y^{-\frac{2}{3}}\,dy} = 6\int{\frac{\sin{x}}{(2 - \cos{x})^3}\,dx}.

    Can you go from here? Hint: the RHS is the same as Q1...
    so

    \int{y^{-\frac{2}{3}}\,\frac{dy}{dx}\,dx} = \int{\frac{6\sin{x}}{(2 - \cos{x})^3}\,dx} =  - \frac{1}{2(2 - cos x)^2} + c
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  14. #14
    ADY
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    This is getting quite a messy thread - only because of my posts, so i'll do a consolidation from post 1.

    Using:

    \int{f\left(u(x)\right)\,\frac{du}{dx}\,dx} = \int{f(u)\,du}

    show that:

    \int{\frac{\sin{x}}{(2 - \cos{x})^3}\,dx} = \int{(2 - \cos{x})^{-3}\sin{x}\,dx}

    where c is an arbitray constant.

    We Let u = 2 - \cos{x} so that \frac{du}{dx} = \sin{x}

    So The integral becomes

    \int{u^{-3}\,\frac{du}{dx}\,dx} = \int{u^{-3}\,du}

     = -\frac{1}{2}u^{-2} + C

     = -\frac{1}{2}(2 - \cos{x})^{-2} + C

     = -\frac{1}{2(2 - \cos{x})^2} + C.[/quote]


    Now in implicit form, find the solution

    \frac{dy}{dx} = \frac{6y^{2/3}sin x}{(2- cos x)^3}

    \frac{dy}{dx} = \frac{6y^{\frac{2}{3}} \sin x}{(2-\cos x)^3 }

    \frac{dy}{6y^{\frac{2}{3}}} = \frac{\sin x\cdot dx}{(2-\cos x)^3}

    \int \frac{dy}{6y^{\frac{2}{3}}} = \int \frac{\sin x\cdot dx}{(2-\cos x)^3}

    In the right side we find it in the 1st part of the question

     \frac{y^{\frac{-2}{3} +1}}{6\left(\frac{-2}{3}+1\right)}=- \frac{1}{2(2 - cos x)^2} + c<br />

    \frac{y^{\frac{1}{3}}}{2} = - \frac{1}{2(2 - cos x)^2} + c<br />

    We now need to find the corresponding implicit solution that satisifies intial condition y = 1 when x = 0

    \frac{1}{2} = \frac{-1}{2(2-1)^2}+c

    \frac{1}{2} = \frac{-1}{2} + c \Rightarrow c = 1 so

    \frac{y^{\frac{1}{3}}}{2} = \frac{-1}{2(2-\cos x)^2} + 1

    y^{\frac{1}{3}} = \frac{-1}{(2-\cos x)^2} + 2


    So to find the explicit form of the solution:

    y = \left( \frac{-1}{(2-\cos x)^2} +2 \right)^3


    Then the value of y when x= \frac{1}{2\pi}

     y = \left( \frac{-1}{(2-\cos (1/2\pi))^2} +2 \right)^3

    = 5.3594
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  15. #15
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    Quote Originally Posted by ADY View Post
    so

    \int{y^{-\frac{2}{3}}\,\frac{dy}{dx}\,dx} = \int{\frac{6\sin{x}}{(2 - \cos{x})^3}\,dx} =  - \frac{1}{2(2 - cos x)^2} + c
    No.

    Perform the integration on the LHS.

    Perform the integration on the RHS.
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