Anonymous
Anonymous asked in Science & MathematicsMathematics · 1 month ago

Having trouble finding the integral here. ?

Having a tough time with this problem. Can’t even figure out the right way to use the hint. Lol

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    e^(x) = u

    e^(x) * dx = du

    5 * e^(x) * dx / sqrt(4 - 9 * e^(2x)) =>

    5 * du / sqrt(4 - 9 * u^2)

    Recall that 1 - sin(t)^2 = cos(t)^2

    4 - 9 * u^2 = 4 - 4 * sin(t)^2

    -9 * u^2 = -4 * sin(t)^2

    u^2 = (4/9) * sin(t)^2

    u = (2/3) * sin(t)

    du = (2/3) * cos(t) * dt

    5 * du / sqrt(4 - 9 * u^2) =>

    5 * (2/3) * cos(t) * dt / sqrt(4 - 4 * sin(t)^2) =>

    (10/3) * cos(t) * dt / sqrt(4 * cos(t)^2) =>

    (10/3) * cos(t) * dt / (2 * cos(t)) =>

    (5/3) * dt

    Integrate

    (5/3) * t + C

    u = (2/3) * sin(t)

    (3/2) * u = sin(t)

    arcsin(3u / 2) = t

    (5/3) * arcsin(3u / 2) + C

    u = e^(x)

    (5/3) * arcsin(3 * e^(x) / 2) + C

    Truth be told, we could have used one substitution, but I find it's oftentimes easier to express an expression in terms of a variable and then express that variable in terms of another expression instead of substituting one expression for another.  It's less of a conceptual leap.

    4 - 9 * e^(2x) = 4 - 4 * sin(t)^2

    9 * e^(2x) = 4 * sin(t)^2

    3 * e^(x) = 2 * sin(t)

    e^(x) = (2/3) * sin(t)

    e^(x) * dx = (2/3) * cos(t) * dt

    5 * e^(x) * dx / sqrt(4 - 9 * e^(2x)) =>

    5 * (2/3) * cos(t) * dt / sqrt(4 - 4 * sin(t)^2) =>

    (10/3) * cos(t) * dt / (2 * cos(t)) =>

    (5/3) * dt

    (5/3) * t + C

    e^(x) = (2/3) * sin(t)

    (3 * e^(x) / 2) = sin(t)

    t = arcsin(3 * e^(x) / 2)

    (5/3) * arcsin(3 * e^(x) / 2) + C

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  • Ian H
    Lv 7
    1 month ago

    d/dy[arcsin(y/2)] = 1/√[4 – y^2] let y = (3/2)e^x 

    d/dx{arcsin[(3/2)e^x]} = 3e^x/√[4 – (3e^x)^2] 

    I = ∫[5e^x/√[4 – 9e^(2x)]dx = (5/3)arcsin[(3/2)e^x] 

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