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how to take the integral of cos(square root of x) dx.
need to use substitution and integration by parts but I don't know what to substitute

  • calculus -

    let t = x^.5
    dt = .5 x^-.5 dx
    so dx = 2 x^.5 dt which is dx = 2 t dt

    so integral of 2 t cos t dt

    that might help

  • calculus -

    (Integral of) cos x^(1/2) dx
    Do the substitution part first.
    Let x^(1/2) = y
    x = y^2
    dx = 2y du
    So the integral becomes, after substitution
    2*(Integral of)y cos y dy
    Now use integration by parts. You may have to use it twice.
    You should end up with
    2 cos y + y sin y
    Then substitute back x in terms of y
    = 2 cos(sqrt x) + 2sqrt x * sin(sqrt x)

  • calculus -

    integral of 2 t cos t dt

    let u = 2 t
    so du = 2 dt
    let dv = cos t dt
    so v = sin t

    u v = 2 t sin t

    v du = 2 sin t dt

    u v - integral v du
    = 2 t sin t + 2 cos t

  • calculus -

    Remarkable :) LOL

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