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Math Help - is this integrer solved properly?

  1. #1
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    Question is this integrer solved properly?

    Let's consider the integrer:

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

    Let's assume that:

    cos(x)+sin(x) = u
     \Big (cos(x)-sin(x) \Big  ) dx = du
     \Bigg  ( \Big (cos(x)-sin(x) \Big  ) \frac{cos(x)+sin(x)}{cos(x)+sin(x)} \Bigg )dx = du
     \frac{cos^{2}(x)-sin^{2}(x)}{cos(x)+sin(x)} dx = du
    \frac{cos(2x)}{u} dx = du
     cos(2x) dx = u du

    then our interger:

    - \int ue^udu

    next:

    f(x) = e^u, f'(x) = e^u
    oraz
    g(x) = u, g'(x) = 1

    and finally we get:

    - \int ue^udu = ue^u - \int e^u = ue^u - e^u +C = e^u(u - 1) + C

    so:

    - \int cos(2x)e^{cos(x)+sin(x)}  = e^{cos(x)+sin(x)} \Big (cos(x)+sin(x)+1 \Big ) + C

    ...and I am not sure if that solution is correct.
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  2. #2
    MHF Contributor Amer's Avatar
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    Jordan
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    -\int cos(2x)e^{cos(x)+sin(x)}=e^{cos(x)+sin(x)} \Big (cos(x)+sin(x)+1 \Big) +c " alt="-\int cos(2x)e^{cos(x)+sin(x)}=e^{cos(x)+sin(x)} \Big (cos(x)+sin(x)+1 \Big) +c " />

    you should just change the red signs negative to positive and the positive to the negative

    you answer is correct

    and this come from


    sinx derive is cos and cosx derive -sinx


    look carefully to this


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  3. #3
    MHF Contributor Amer's Avatar
    Joined
    May 2009
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    Jordan
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    1,093
    <br /> <br />
((-))\int {cos(2x)e^{cos(x)+sin(x)}}=e^{cos(x)+sin(x)} \Big (cos(x)+sin(x)((+))1 \Big) +c <br /> <br />


    you should just change the sign in the arcs negative to positive and the positive to the negative

    you answer is correct

    and this come from


    sinx derive is cos and cosx derive -sinx


    look carefully to this


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