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Math Help - how would I Evaluate this??

  1. #1
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    how would I Evaluate this??

    ∫[(sinxdx)/(cos^2x + cos x -2)]


    what i did was integration by parts but it is wrong.....and i couldn't even finish it

    =∫sinxdx/(cosx-1)(cosx+2)


    sinx/cos^2+cosx-2 = A/cosx-1 + B/cosx +2
    sinx = Acosx + 2A + Bcosx -B
    sinx = cosx(A+B) + (2A-B)

    2A-B=0
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  2. #2
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    Quote Originally Posted by Yello View Post
    ∫[(sinxdx)/(cos^2x + cos x -2)]


    what i did was integration by parts but it is wrong.....and i couldn't even finish it

    =∫sinxdx/(cosx-1)(cosx+2)


    sinx/cos^2+cosx-2 = A/cosx-1 + B/cosx +2
    sinx = Acosx + 2A + Bcosx -B
    sinx = cosx(A+B) + (2A-B)

    2A-B=0
    \int{\frac{\sin{x}}{\cos^2{x} + \cos{x} - 2}\,dx}.

    You need to use a u substitution:

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


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

     = -\int{\left(\frac{1}{u^2 + u - 2}\right)\frac{du}{dx}\,dx}

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


    Now use a partial fractions decomposition.
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