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Math Help - Deriving some inverse trig functions and some integrals

  1. #1
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    Deriving some inverse trig functions and some integrals

    Hi all!

    I am having some trouble with a few problems I've been working on.

    Find the derivatives of the function

     y = x^3arctan(1-x) + arcsin(2x)

    Step was was recognizing the product rule for  x^3acrtan(1-x)

    so I have  3x^2arctan(1-x) + x^3(-1/(\sqrt{1-(-1)^2)} + 2/(\sqrt{1-(2x)^2})

    but I am not sure that it is right



    Problem number two: Find the derivative of

      y = tan(arccos(x)) + arccsc(3x^2)

    this one threw me a monkey wrench right away since the argument to arccos is only x and not a function of x. even if it were more then just x i'm still scratching my head on how to start this one

    Problem three:

    Intergrate

     /int 4 / (8 + 9x^2 -6x)

    I figured to used the identity that  \int du / (a^2 + u^2) = (1 / a)arctan(u/a) + C

    So I then went to complete the square to give me what I wanted

    I got  [(3x)^2 + 2(3x) + 1^2 ] -1

    so then my denominator turned into 7 + (3x+1)^2 which I figured was wrong. A little confused on this one.


    Same for this last problem

     \int 5/(\sqrt{4-x^2+2x})


    any help would be appreciated!!!
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  2. #2
    MHF Contributor Calculus26's Avatar
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    1) d/dx(arctan(1-x) = -1/(1-(1-x)^2 = -1/sqrt(2x-x^2)


    2) tan(arcos(x)) = sqrt(1-x^2)/x draw a right triangle with angle whose cosine is x

    dsqrt(1-x^2)/x /dx = -1/sqrt(1-x^2) - sqrt(1-x^2)/x^2

    = -1/[x^2 sqrt(1-x^2)]

    Even without this d(tan(u)/dx = sec^2 (u)8du/dx

    d(arctan(cos(x))/dx = sec^2 (cos(x)) [-1/sqrt(1-x^2)]


    sec(arccos(x)) = 1/x = -1/[x^2 sqrt(1-x^2)]


    for 3) 9x^2 -6x +8 = (3x-1)^2 + 7 let u =3x-1 and note a = sqrt(7)
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