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

• Sep 7th 2010, 03:50 PM
tydube
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)
• Sep 7th 2010, 04:53 PM
skeeter
cleared the fraction in the radical to make life easier ...

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

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

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

$u = x^2$

$du = 2x \, dx$

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

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

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

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

$\displaystyle -k \int x^2 \cdot \frac{-x}{\sqrt{k^2-x^2}} \, dx = -k\left[x^2\sqrt{k^2-x^2} + \frac{2}{3}(k^2-x^2)^{\frac{3}{2}} + C\right]$
• Sep 7th 2010, 06:37 PM
tydube
thanks
(Nod)
Quote:

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

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

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

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

$u = x^2$

$du = 2x \, dx$

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

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

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

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

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

• Sep 7th 2010, 09:17 PM
TheCoffeeMachine
Without 'by-parts'
Rewrite the integral as following:

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

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

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

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

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

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

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

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

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

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

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

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

$\displaystyle \int\frac{x^3}{\sqrt{1-\frac{x}{k^2}}}\;{dx} = -\frac{k}{3}\left[3k^2\sqrt{k^2-x^2}-(k^2-x^2)^{\frac{3}{2}}\right]+C$.
• Sep 7th 2010, 09:44 PM
sa-ri-ga-ma
By substitution method.

Given integration can be written as

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

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

.
• Sep 9th 2010, 06:44 PM
tydube
thanks
(Nod)
Quote:

Originally Posted by TheCoffeeMachine
Rewrite the integral as following:

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

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

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

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

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

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

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

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

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

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

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

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

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