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Thread: Integrate [x^3/sqrt(1-x^2/k^2)]dx

  1. #1
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    Integrate [x^3/sqrt(1-x^2/k^2)]dx

    Can someone please check my process, also please can you advise on an easier way to write mathematical notation on this forum?

    Integrate ( x^3/(1-x^2/k^2))dx
    where k is a constant.
    Let u = x^2, dv = x/sqrt(1-x^2/k^2)

    Integration by parts.

    Integral fx = uv - integral vdu
    v = -k^2*sqrt(1-x^2/k^2)
    du = 2xdx


    susbstituting for u,v,du

    Integral fx = -k^2 *x^2*sqrt(1-x^2/k^2)+ integral [2xk^2*sqrt(1-x^2/k^2)dx]
    Integral fx = -k^2 *x^2*sqrt(1-x^2/k^2) - 2/3*k^4*((1-x^2/k^2)^3/2)
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  2. #2
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    cleared the fraction in the radical to make life easier ...

    $\displaystyle \displaystyle \frac{k}{k} \cdot \frac{x^3}{\sqrt{1 - \frac{x^2}{k^2}}}$

    $\displaystyle \displaystyle \frac{kx^3}{\sqrt{k^2\left(1 - \frac{x^2}{k^2}\right)}}
    $



    $\displaystyle \displaystyle -k \int x^2 \cdot \frac{-x}{\sqrt{k^2-x^2}} \, dx$

    $\displaystyle u = x^2$

    $\displaystyle du = 2x \, dx$

    $\displaystyle dv = \frac{-x}{\sqrt{k^2-x^2}} \, dx$

    $\displaystyle v = \sqrt{k^2-x^2}$


    $\displaystyle \displaystyle -k \int x^2 \cdot \frac{-x}{\sqrt{k^2-x^2}} \, dx = -k\left[x^2\sqrt{k^2-x^2} - \int 2x\sqrt{k^2-x^2} \, dx\right]$

    $\displaystyle \displaystyle -k \int x^2 \cdot \frac{-x}{\sqrt{k^2-x^2}} \, dx = -k\left[x^2\sqrt{k^2-x^2} + \int -2x\sqrt{k^2-x^2} \, dx\right]$

    $\displaystyle \displaystyle -k \int x^2 \cdot \frac{-x}{\sqrt{k^2-x^2}} \, dx = -k\left[x^2\sqrt{k^2-x^2} + \frac{2}{3}(k^2-x^2)^{\frac{3}{2}} + C\right]$
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  3. #3
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    thanks

    Quote Originally Posted by skeeter View Post
    cleared the fraction in the radical to make life easier ...

    $\displaystyle \displaystyle \frac{k}{k} \cdot \frac{x^3}{\sqrt{1 - \frac{x^2}{k^2}}}$

    $\displaystyle \displaystyle \frac{kx^3}{\sqrt{k^2\left(1 - \frac{x^2}{k^2}\right)}}
    $



    $\displaystyle \displaystyle -k \int x^2 \cdot \frac{-x}{\sqrt{k^2-x^2}} \, dx$

    $\displaystyle u = x^2$

    $\displaystyle du = 2x \, dx$

    $\displaystyle dv = \frac{-x}{\sqrt{k^2-x^2}} \, dx$

    $\displaystyle v = \sqrt{k^2-x^2}$


    $\displaystyle \displaystyle -k \int x^2 \cdot \frac{-x}{\sqrt{k^2-x^2}} \, dx = -k\left[x^2\sqrt{k^2-x^2} - \int 2x\sqrt{k^2-x^2} \, dx\right]$

    $\displaystyle \displaystyle -k \int x^2 \cdot \frac{-x}{\sqrt{k^2-x^2}} \, dx = -k\left[x^2\sqrt{k^2-x^2} + \int -2x\sqrt{k^2-x^2} \, dx\right]$

    $\displaystyle \displaystyle -k \int x^2 \cdot \frac{-x}{\sqrt{k^2-x^2}} \, dx = -k\left[x^2\sqrt{k^2-x^2} + \frac{2}{3}(k^2-x^2)^{\frac{3}{2}} + C\right]$
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  4. #4
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    Without 'by-parts'

    Rewrite the integral as following:

    $\displaystyle \displaystyle \int\frac{x^3}{\sqrt{1-\frac{x}{k^2}}}\;{dx}$ $\displaystyle \displaystyle = \int\frac{x^3}{\sqrt{\frac{1}{k^2}(k^2-x^2)}}}\;{dx} = $ $\displaystyle \displaystyle = \int\frac{x^3}{\frac{1}{k}\sqrt{k^2-x^2}}}\;{dx} = $ $\displaystyle \displaystyle = \int\frac{kx^3}{\sqrt{k^2-x^2}}}\;{dx} $

    Let $\displaystyle t = \sqrt{k^2-x^2}$ and (using the chain rule) differentiate this with respect to $\displaystyle x$:

    $\displaystyle \displaystyle \frac{dt}{dx} = \left(\sqrt{k^2-x^2\right)'$ $\displaystyle \displaystyle= \frac{\left(k^2-x^2\right)'}{2\sqrt{k^2-x^2}}$ $\displaystyle \displaystyle= \frac{-2x}{2\sqrt{k^2-x^2}}$ $\displaystyle \displaystyle= \frac{-x}{\sqrt{k^2-x^2}}$

    Solving this for $\displaystyle dx$, as our purpose was, we have:

    $\displaystyle \displaystyle dx = -\left(\frac{\sqrt{k^2-x^2}}{x}\right){dt}$

    Going back to our original integral and putting that in for $\displaystyle dx$ we get:

    $\displaystyle \displaystyle \int\frac{kx^3}{\sqrt{k^2-x^2}}}\;{dx}$ $\displaystyle \displaystyle = -\int\left(\frac{kx^3}{\sqrt{k^2-x^2}}}\right)\;\left(\frac{\sqrt{k^2-x^2}}{x}\right){dt}}$ $\displaystyle \displaystyle = -\int kx^2 \;{dt}$

    Finding $\displaystyle x^2$ in terms of $\displaystyle t$ from the relation of our substitution gives us:

    $\displaystyle t = \sqrt{k^2-x^2} \Rightarrow t^2 = k^2-x^2 \Rightarrow t^2-k^2 = -x^2 \Rightarrow k^2-t^2 = x^2$.

    Putting that in for $\displaystyle x^2$, we get a simple function in the integrand that is easy to integrate:

    $\displaystyle \displaystyle -\int kx^2 \;{dt}$ $\displaystyle \displaystyle = -\int k(k^2-t^2) \;{dt}$ $\displaystyle \displaystyle = -k\left(k^2t-\frac{t^3}{3}\right)+C = -\frac{k}{3}\left(3k^2t-t^3\right)+C$

    Substituting back for what we have let $\displaystyle t$ to be, we finally get:

    $\displaystyle \displaystyle \int\frac{x^3}{\sqrt{1-\frac{x}{k^2}}}\;{dx} = -\frac{k}{3}\left[3k^2\sqrt{k^2-x^2}-(k^2-x^2)^{\frac{3}{2}}\right]+C$.
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  5. #5
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    By substitution method.

    Given integration can be written as

    $\displaystyle \int{\frac{kx^3}{\sqrt{k^2 -x^2}}}\;{dx}$

    Let x = ksinθ. dx = kcosθdθ. k^2 - x^2 = k^2cos^2θ. Substitute these values in the integration and simplify.

    .
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  6. #6
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    thanks

    Quote Originally Posted by TheCoffeeMachine View Post
    Rewrite the integral as following:

    $\displaystyle \displaystyle \int\frac{x^3}{\sqrt{1-\frac{x}{k^2}}}\;{dx}$ $\displaystyle \displaystyle = \int\frac{x^3}{\sqrt{\frac{1}{k^2}(k^2-x^2)}}}\;{dx} = $ $\displaystyle \displaystyle = \int\frac{x^3}{\frac{1}{k}\sqrt{k^2-x^2}}}\;{dx} = $ $\displaystyle \displaystyle = \int\frac{kx^3}{\sqrt{k^2-x^2}}}\;{dx} $

    Let $\displaystyle t = \sqrt{k^2-x^2}$ and (using the chain rule) differentiate this with respect to $\displaystyle x$:

    $\displaystyle \displaystyle \frac{dt}{dx} = \left(\sqrt{k^2-x^2\right)'$ $\displaystyle \displaystyle= \frac{\left(k^2-x^2\right)'}{2\sqrt{k^2-x^2}}$ $\displaystyle \displaystyle= \frac{-2x}{2\sqrt{k^2-x^2}}$ $\displaystyle \displaystyle= \frac{-x}{\sqrt{k^2-x^2}}$

    Solving this for $\displaystyle dx$, as our purpose was, we have:

    $\displaystyle \displaystyle dx = -\left(\frac{\sqrt{k^2-x^2}}{x}\right){dt}$

    Going back to our original integral and putting that in for $\displaystyle dx$ we get:

    $\displaystyle \displaystyle \int\frac{kx^3}{\sqrt{k^2-x^2}}}\;{dx}$ $\displaystyle \displaystyle = -\int\left(\frac{kx^3}{\sqrt{k^2-x^2}}}\right)\;\left(\frac{\sqrt{k^2-x^2}}{x}\right){dt}}$ $\displaystyle \displaystyle = -\int kx^2 \;{dt}$

    Finding $\displaystyle x^2$ in terms of $\displaystyle t$ from the relation of our substitution gives us:

    $\displaystyle t = \sqrt{k^2-x^2} \Rightarrow t^2 = k^2-x^2 \Rightarrow t^2-k^2 = -x^2 \Rightarrow k^2-t^2 = x^2$.

    Putting that in for $\displaystyle x^2$, we get a simple function in the integrand that is easy to integrate:

    $\displaystyle \displaystyle -\int kx^2 \;{dt}$ $\displaystyle \displaystyle = -\int k(k^2-t^2) \;{dt}$ $\displaystyle \displaystyle = -k\left(k^2t-\frac{t^3}{3}\right)+C = -\frac{k}{3}\left(3k^2t-t^3\right)+C$

    Substituting back for what we have let $\displaystyle t$ to be, we finally get:

    $\displaystyle \displaystyle \int\frac{x^3}{\sqrt{1-\frac{x}{k^2}}}\;{dx} = -\frac{k}{3}\left[3k^2\sqrt{k^2-x^2}-(k^2-x^2)^{\frac{3}{2}}\right]+C$.
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