calculus 2

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how to find the integral of (3x+pi)cos3xdx from [pi over 6, pi over 3}

  • calculus 2 -

    integrate by parts
    int u dv = u v - int v du

    dv = cos 3 x dx
    so
    v = (1/3) sin 3x

    u = 3 x + pi
    du = 3 dx

    so
    u v - (3/3)integral of sin 3x dx
    put in your limits
    (3x+pi)(sin 3x)- int sin3x dx
    at pi/3 minus at pi/6

  • calculus 2 -

    i really appreciate the work it helps me a lot thank you

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