calculus

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INTEGRATION BY PARTS...

1) e^5t sin 3t dt
2) xe^3x dx
3) x sec^2 x dx

**anybody can help me for this question?

  • calculus -

    if we let A be our integral, we have

    A = ∫e^5t sin3t dt
    u=e^5t
    du = 5e^5t dt

    dv = sin3t dt
    v = -1/3 cos3t

    A = uv - ∫v du
    = -1/3 e^5t cos3t + 5/3 ∫e^5t cos3t dt
    u=e^5t
    du = 5e^5t dt
    dv = cos3t dt
    v = 1/3 sin3t

    A = -1/3 e^5t cos3t + 5/3(1/3 e^5t sin3t - 5/3 ∫e^5t sin3t dt)
    = -1/3 e^5t cos3t + 5/9 e^5t sin3t - 25/9 A
    34/9 A = 1/9 e^5t (-3cos3t + 5sin3t)
    A = 1/34 e^5t (-3cos3t + 5sin3t)
    -----------------------------------
    A = ∫xe^3x dx
    u = x
    du = dx
    dv = e^3x dx
    v = 1/3 e^3x

    A = uv - ∫v du
    = 1/3 xe^3x - 1/3 ∫e^3x dx
    = 1/3 xe^3x - 1/9 e^3x
    = 1/9 e^3x (3x-1)
    ------------------------
    A = ∫x sec^2 x dx
    u = x
    du = dx
    dv = sec^2 x dx
    v = tan x

    A = uv - ∫v du
    = x tanx - ∫tanx dx
    = x tanx - ∫sinx/cosx dx
    = x tanx + log cosx

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