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Math Help - int x^2sin^-1(x)

  1. #1
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    int x^2sin^-1(x)

    \int x^2\sin^{-1}{x}
    let dv = x^2
    v = x^3/3
    let u = sin^-1{x}
    du = \frac{1}{(\sqrt{1 + x^2}} dx

    \frac{x^3}{3}\sin^{-1}{x} - \int\frac{x^3}{3}\frac{1}{(\sqrt{1-x^2})}

    for the 2nd integral:
    dv = \frac{x}{\sqrt{1-x^2}}dx
    v = - \sqrt{1-x^2}

    u = 2x
    du = 2dx

    -x^2\sqrt{1-x^2} - \int -2x\sqrt{1-x^2}dx
    integrat

    and my answer wud be:

    \frac{x^3}{3}\sin^{-1}{x} +\frac{1}{3}x^2\sqrt{1-x^2} +\frac{2}{9}(1-x^2)^{\frac{3}{2}}

    but the answer at the back of book is:
    \frac{1}{3}x^3\sin^{-1}{x} + \frac{1}{9}(x^2+2)\sqrt{1-x^2} + C
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  2. #2
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    \int x^2\sin^{-1}{x}
    let dv = x^2
    v = x^3/3
    let u = sin^-1{x}
    du = \frac{1}{(\sqrt{1 + x^2}} dx

    \frac{x^3}{3}\sin^{-1}{x} - \int\frac{x^3}{3}\frac{1}{(\sqrt{1-x^2})}dx
    ......
    for the 2nd integral
    For the 2nd integral let x=sint.
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  3. #3
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    using trigo sub
    i will get a value of
    \frac{1}{3}\sin^{-1}x + \frac{1}{3}\sqrt{1-x^2} - \frac{1}{9}(1-x^2)^{\frac{3}{2}}

    how do i simplify to get the corect answer?
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  4. #4
    Math Engineering Student
    Krizalid's Avatar
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    \frac{x^3}{3}\sin^{-1}{x} - \int\frac{x^3}{3}\frac{1}{(\sqrt{1-x^2})}
    For the second one set u=\sqrt{1-x^2}
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  5. #5
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    using trigo sub
    i will get a value of
    \frac{1}{3}\sin^{-1}x + \frac{1}{3}\sqrt{1-x^2} - \frac{1}{9}(1-x^2)^{\frac{3}{2}}

    how do i simplify to get the corect answer?
    The correct answer is
    \frac{1}{3}x^3\sin^{-1}x + \frac{1}{3}\sqrt{1-x^2} - \frac{1}{9}(1-x^2)^{\frac{3}{2}}
    where
    Attached Thumbnails Attached Thumbnails int x^2sin^-1(x)-aug7.gif  
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