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Math Help - Big problem with integral

  1. #1
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    Big problem with integral

    \int\frac{sinx}{cosx-2sinx+2}dx
    I found the following relations:
    tan\frac{x}{2}=t
    sinx=\frac{2t}{1+t^2}
    cosx=\frac{1-t^2}{1+t^2}
    dx=\frac{dt}{1+t^2}
    Then I did substitutions and got:
    \int\frac{(1-t^2)+2(1+t^2)-4t}{2t(1+t^2)}dt=\int\frac{t^2-4t+3}{2t(1+t^2)}
    Then:
    t^2-4t+3=A(1+t^2)+(Mt+N)2t
    Got that:
    A=3
    M=-1
    N=4
    Thus:
    \int\frac{t^2-4t+3}{2t(1+t^2)}dt=\int\frac{A}{2t}+\int\frac{Mt+N  }{1+t^2}=\frac{3}{2}\int\frac{dt}{t}-\int\frac{tdt}{1+t^2}+4\int\frac{dt}{1+t^2}

    Am I very lost ?
    It doesn't seem to work out according to math book answer
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  2. #2
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    Hello, totalnewbie!

    Your substitution is off . . .


    \int\frac{\sin x\,dx}{\cos x-2\sin x+2}

    I found the following relations:
    \tan\frac{x}{2}=t\qquad \sin x=\frac{2t}{1+t^2} \qquad \cos x=\frac{1-t^2}{1+t^2} \qquad dx = \frac{dt}{1+t^2}

    Substitute: . \int\frac{\frac{2t}{1+t^2}\cdot\frac{2\,dt}{1+t^2}  }{\frac{1-t^2}{1+t^2} - \frac{4t}{1+t^2} + 2} = \;\int\frac{\frac{4t\,dt}{(1+t^2)^2}}{\frac{t^2 - 4t + 3}{1 + t^2}} = \;\int\frac{4t\,dt}{(t-1)(t-3)(t^2+1)}


    Use partial fractions: . \frac{4t}{(t-1)(t-3)(t^2+1)}\;=\;\frac{A}{t-1} + \frac{B}{t-3} + \frac{Ct + D}{t^2+1}

    . . I got: . A = -1,\;B = \frac{3}{5},\;C = \frac{2}{5},\;D = -\frac{4}{5}


    Integrate: . -\!\int\frac{dt}{t-1} \;+\;\frac{3}{5}\!\int\frac{dt}{t-3} \;+ \;\frac{2}{5}\!\int\frac{t\,dt}{t^2+1} \;- \;\frac{4}{5}\!\int\frac{dt}{t^2+1}
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  3. #3
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    Quote Originally Posted by Soroban View Post
    Hello, totalnewbie!

    Your substitution is off . . .



    Substitute: . \int\frac{\frac{2t}{1+t^2}\cdot\frac{2\,dt}{1+t^2}  }{\frac{1-t^2}{1+t^2} - \frac{4t}{1+t^2} + 2} = \;\int\frac{\frac{4t\,dt}{(1+t^2)^2}}{\frac{t^2 - 4t + 3}{1 + t^2}} = \;\int\frac{4t\,dt}{(t-1)(t-3)(t^2+1)}


    Use partial fractions: . \frac{4t}{(t-1)(t-3)(t^2+1)}\;=\;\frac{A}{t-1} + \frac{B}{t-3} + \frac{Ct + D}{t^2+1}

    . . I got: . A = -1,\;B = \frac{3}{5},\;C = \frac{2}{5},\;D = -\frac{4}{5}


    Integrate: . -\!\int\frac{dt}{t-1} \;+\;\frac{3}{5}\!\int\frac{dt}{t-3} \;+ \;\frac{2}{5}\!\int\frac{t\,dt}{t^2+1} \;- \;\frac{4}{5}\!\int\frac{dt}{t^2+1}
    Something is wrong with constants.
    Book gives the following constants:-1, 3/5, -2/5, 1/5
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  4. #4
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    Hello, totalnewbie!

    Something is wrong with constants.
    Book gives the following constants: -1, 3/5, -2/5, 1/5

    Then something is wrong with the book.

    My constants check out . . . Theirs don't!

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  5. #5
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    Quote Originally Posted by Soroban View Post
    Hello, totalnewbie!


    Then something is wrong with the book.

    My constants check out . . . Theirs don't!

    You have got 4tdt. It must be 2tdt.

    And checking out if we have 4t instead of 2t...
    We have 4t=A(t-3)(t*t+1)+B(t-1)(t*t+1)+(Ct+D)(t-1)(t-3)
    Substitute your A, B, C and D.
    And then take t as 2, you can see that they are not equal.
    If you take t as 100, they are equal.

    I dont get the point.
    Last edited by totalnewbie; December 15th 2006 at 02:29 PM.
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  6. #6
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    I really need help with this excercise.
    Please help anybody.
    Last edited by totalnewbie; December 16th 2006 at 04:51 PM.
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  7. #7
    Grand Panjandrum
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    Quote Originally Posted by totalnewbie View Post
    \int\frac{sinx}{cosx-2sinx+2}dx
    I found the following relations:
    \tan\frac{x}{2}=t
    \sin x=\frac{2t}{1+t^2}
    \cos x=\frac{1-t^2}{1+t^2}
    dx=\frac{dt}{1+t^2}
    The above all checks out. Now the substitution and simplification is a bit
    to complex to be done reliably so doing it mechanicaly what you should end
    up with is shown in the attachment.

    This is what soroban gets for this, but with a factor of 2 rather than 4.

    The partial fraction expansion is also shown in the attachment, but this
    is again different from your book

    RonL
    Attached Thumbnails Attached Thumbnails Big problem with integral-gash.jpg  
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