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Math Help - Integral with substitution and parts

  1. #1
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    Integral with substitution and parts

    ∫ sin(2t)*e^[cos(t)] dt (from t=0 to π)

    My work:
    trig identity: sin2t= 2sintcost
    bring constant to the outside
    => 2 ∫ sin(t)cos(t)*e^[cos(t)] dt (from t=0 to π)
    I did u substitution and did
    u=cos(t)
    du=-sin(t)dt

    new intervals:
    cos(pi)=-1
    cos(0)=1

    Im not sure what to after this. Do I divide du=-sint(dt) by negative 1 and bring that -1 to the outside of integral since there is only a sin(t)dt not -sin(t)dt?

    I appreciate any help!

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  2. #2
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    Re: Integral with substitution and parts

    Ok I figured out what to do, but this last algebra is bugging me.

    = 2 * [(e - e) - (-1/e - 1/e)]
    = 4/e

    does it simplify to 0 (+2/e +2/e)

    which equals 4/e?
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  3. #3
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    Re: Integral with substitution and parts

    Hello, Steelers72!

    I couldn't follow what you did, but you got the right answer.


    \int^{\pi}_0\sin(2t)e^{\cos t}\,dt

    I'll drop the limits for now . . .

    We have: . I \;=\;\int 2\sin t\cos t\,e^{\cos t}dt \;=\;2\int\cos t(\sin t\,e^{\cos t}dt)

    By parts: . \begin{Bmatrix}u &=& \cos t && dv &=& \sin t\,e^{\cos t}dt \\ du &=& \text{-}\sin t\,dt && v &=& \text{-}e^{\cos t} \end{Bmatrix}

    Then: . I \;=\;2\left[\text{-}e^{\cos t}\cos t - \int e^{\cos t}\sin t\,dt\right]

    . . . . . I \;=\;2\left[\text{-}e^{\cos t}\cos t + \int e^{\cos t}(\text{-}\sin t\,dt)\right]

    . . . . . I \;=\;2\left[\text{-}e^{\cos t}\cos t + e^{\cos t}\right] + C

    . . . . . I \;=\;2e^{\cos t}(1-\cos t) + C


    Evaluate: . 2e^{\cos t}(1-\cos t)\,\bigg]^{\pi}_0

    . . . . . . . =\;2e^{\cos\pi}(1-\cos\pi) - 2e^{\cos0}(1 - \cos 0)

    . . . . . . . =\;2e^{-1}(1+1) - 2e^1(1-1)

    . . . . . . . =\;\frac{4}{e}
    Thanks from Steelers72
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  4. #4
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    Re: Integral with substitution and parts

    Thanks for the help! Just one question:

    When you took the integral of e^cost (-sintdt) , why did the -sint get eliminated? The antiderivative of -sin is cos right? I'm confused at that part
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  5. #5
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    Re: Integral with substitution and parts

    Alternatively making a substitution can make the integral simple.
    let cos t = u
    -sint dt = du
    thus we will get


    Integral with substitution and parts-inte-10-mar.png
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