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Math Help - Integration by parts

  1. #1
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    Integration by parts

    Integrate (x^3 dx) / square root of (1-x^2)

    i have to integrate it by parts. i already tried it but my answer waa 0=0 is that possible?
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  2. #2
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    how?

    Quote Originally Posted by confusedgirl View Post
    Integrate (x^3 dx) / square root of (1-x^2)

    i have to integrate it by parts. i already tried it but my answer waa 0=0 is that possible?
    Curious how did you end up with 0?
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  3. #3
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    let u = x^3
    du = 3x^2 dx0
    dv = dx / sq rt of (1-x^2)
    v= arcsin x
    = x^3 arcsin x - integral of 3x^2 arcsinx dx
    u = arcsin x
    du = dx / sq rt of (1-x^2)
    dv = 3x^2 dx
    v = x^3
    = x^3 arcsin x - x^3 arcsin x + integral of x^3 dx / sq rt (1-x^2)
    0=0
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  4. #4
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    first find integrate:sin^-1 (x)
    then go on integrating keeping x^3 as the part to differentiate everytime.
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  5. #5
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    first find integrate:sin^-1 (x)
    then go on integrating keeping x^3 as the part to differentiate everytime.
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  6. #6
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    Hello, confusedgirl!

    Watch very carefully . . .


    I \;=\;\int\frac{x^3\,dx}{\sqrt{1-x^2}}

    We have: . I \;=\;\int(x^2)\left[x(1-x^2)^{-\frac{1}{2}}dx\right]

    \text{By parts: }\;\begin{array}{ccccccc}u &=& x^2 && dv &=& x(1-x^2)^{-\frac{1}{2}}dx \\<br />
du &=& 2x\,dx && v &=& -(1-x^2)^{\frac{1}{2}} \end{array}

    Then: . I \;=\;-x^2(1-x^2)^{\frac{1}{2}} + \int 2x(1-x^2)^{\frac{1}{2}}dx


    . . For the second integral, use "normal" substitution.

    . . We have: . J \;=\;\int (1-x^2)^{\frac{1}{2}}(2x\,dx)

    . . Let: . u \,=\,1-x^2\quad\Rightarrow\quad du \,=\,-2x\,dx \quad\Rightarrow\quad 2x\,dx \,=\,-du

    . . Substitute: . J \;=\;\int u^{\frac{1}{2}}(-du) \;=\;-\int u^{\frac{1}{2}}\,du \:=\:-\tfrac{2}{3}u^{\frac{3}{2}} + C \;=\;-\tfrac{2}{3}(1-x^2)^{\frac{3}{2}} + C


    Therefore: . I \;=\;-x^2(1-x^2)^{\frac{1}{2}} - \tfrac{2}{3}(1-x^2)^{\frac{3}{2}} + C


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    Although we have completed the integration, your book's answer
    . . is often simplified beyond all recognition.

    Here's how they do it . . .


    We have: . -x^2(1-x^2)^{\frac{1}{2}} - \tfrac{2}{3}(1-x^2)^{\frac{3}{2}}

    Factor: . -\tfrac{1}{3}(1-x^2)^{\frac{1}{2}}\bigg[3x^2 + 2(1-x^2)\bigg] . . . .
    hope you followed that!

    . . . . =\;-\tfrac{1}{3}(1-x^2)^{\frac{1}{2}}\bigg[3x^2 + 2 - 2x^2\bigg]

    . . . . =\;-\tfrac{1}{3}(1-x^2)^{\frac{1}{2}}(x^2+2) . . . . ta-DAA!

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  7. #7
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    hi confusedgirl,

    \int{\frac{x^3}{\sqrt{1-x^2}}}dx

    y=1-x^2\ \Rightarrow\ x^2=1-y

    \frac{dy}{dx}=-2x\ \Rightarrow\ dy=-2xdx\ \Rightarrow\ x^3dx=-\frac{1}{2}x^2dy=\frac{1}{2}(y-1)dy

    \frac{1}{2}\int{\frac{(y-1)}{\sqrt{y}}}dy=\frac{1}{2}\int{\left(\sqrt{y}-\frac{1}{\sqrt{y}}\right)}dy=\frac{1}{2}\int{\left  (y^{\frac{1}{2}}-y^{-\frac{1}{2}}\right)}dy

    This can then be integrated using the power rule.

    To integrate by parts...

    \frac{1}{2}\int{\frac{(y-1)}{\sqrt{y}}}dy

    u=y-1\ du=dy

    dv=y^{-\frac{1}{2}}\ \Rightarrow\ v=\int{y^{-\frac{1}{2}}}dy=2\sqrt{y}

    \frac{1}{2}\int{u}dv=\frac{1}{2}\left(uv-\int{v}du\right)=\frac{1}{2}\left((y-1)2\sqrt{y}-\int{2\sqrt{y}}dy\right)

    \frac{1}{2}\left(-x^2(2)\sqrt{1-x^2}-\frac{4}{3}\left(1-x^2\right)\sqrt{1-x^2}\right)+C

    =-x^2\sqrt{1-x^2}-\frac{2}{3}\left(1-x^2\right)\sqrt{1-x^2}+C=\sqrt{1-x^2}\left(-x^2-\frac{2}{3}+\frac{2}{3}x^2\right)+C

    =-\frac{1}{3}\sqrt{1-x^2}\left(x^2+2\right)+C
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  8. #8
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    thanks
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