# having trouble with this integral

• Jun 18th 2007, 06:59 AM
having trouble with this integral
can someone help using simple techniques of antidifferentiation? / substitution?
i cant get this one
$\int \frac{x^3}{\sqrt{1-2x^2}}dx$
• Jun 18th 2007, 07:37 AM
Pterid
I have one rough solution...

The integral to be solved is:

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

Split it up for convenience:

$\int \frac{x^3}{\sqrt{1-2x^2}} dx = \int \left( \frac{x}{\sqrt{1 - 2x^2}}\right) \left( x^2\right) dx = \int \left( \frac{-2x}{\sqrt{1 - 2x^2}}\right) \left( -\frac{x^2}{2}\right) dx$

...and the first bracket is now the derivative of $\sqrt{1 - 2x^2}$. Hence, the integrand lends itself to integration by parts.

$\int \left( \frac{-2x}{\sqrt{1 - 2x^2}}\right) \left( -\frac{x^2}{2}\right) dx$

$= \left[ \sqrt{1 - 2x^2}\cdot \frac{-x^2}{2}\right] - \int \left(\sqrt{1 - 2x^2}\right) \cdot (-x) dx$

$= \left[ \sqrt{1 - 2x^2}\cdot \frac{-x^2}{2}\right] - \frac{1}{4}\int \left(\sqrt{1 - 2x^2}\right) \cdot (-4x) dx$

$= \left[ \sqrt{1 - 2x^2}\cdot \frac{-x^2}{2}\right] - \frac{1}{4} \cdot \frac{2}{3} \left[ (1 - 2x^2)^{3/2}\right] + C$

$= \left[ \sqrt{1 - 2x^2}\cdot \frac{-x^2}{2}\right] - \frac{1}{6} \left[ \sqrt{1 - 2x^2} (1 - 2x^2)\right] + C$

$= \sqrt{1 - 2x^2}\cdot \left[ - \frac{x^2}{2} - \frac{1}{6} + \frac{x^2}{3} \right] + C$

$= - \frac{1}{6} \sqrt{1 - 2x^2} (x^2 + 1) + C$

This agrees with The Integrator (i.e. Mathematica).
• Jun 18th 2007, 07:50 AM
Plato
$u = 1 - 2x^2 \quad \Rightarrow \quad du = - 4xdx\quad \Rightarrow \quad xdx = - u/4$

$\int {\frac{{x^3 dx}}{{\sqrt {1 - 2x^2 } }} = \int {\frac{{\left( {\frac{{1 - u}}{2}} \right)^2 \left( {\frac{{ - du}}{4}} \right)}}{{\sqrt u }}} }$
• Jun 18th 2007, 10:51 AM
Krizalid
Set $u=\sqrt{1-2x^2}$

The integral becomes to

\begin{aligned}
\int {\frac{{x^3 }}
{{\sqrt {1 - 2x^2 } }}~dx} ~&=~ - \frac{1}
{4}\int {(1 - u^2 )~du}\\
~&=~ - \frac{1}
{4}\left( {u - \frac{1}
{3}u^3 } \right) + k\\
~&=~ - \frac{1}
{6}(x^2 + 1)\sqrt {1 - 2x^2 } + k,~~k\in\mathbb{R}
\end{aligned}
• Jun 18th 2007, 03:38 PM
ty!!!
• Jun 18th 2007, 03:56 PM
Quote:

Originally Posted by Krizalid
Set $u=\sqrt{1-2x^2}$

The integral becomes to

\begin{aligned}
\int {\frac{{x^3 }}
{{\sqrt {1 - 2x^2 } }}~dx} ~&=~ - \frac{1}
{4}\int {(1 - u^2 )~du}\\
~&=~ - \frac{1}
{4}\left( {u - \frac{1}
{3}u^3 } \right) + k\\
~&=~ - \frac{1}
{6}(x^2 + 1)\sqrt {1 - 2x^2 } + k,~~k\in\mathbb{R}
\end{aligned}

hi
was wondering how did you substitute to (1-u^2)?
and this is not part of integration by parts
• Jun 18th 2007, 04:01 PM
Krizalid
If $u=\sqrt{1-2x^2}\implies{du=-\frac{2x}{\sqrt{1-2x^2}}}~dx$

Plus

$u=\sqrt{1-2x^2}\iff{x^2=\frac{1-u^2}2}$
• Jun 19th 2007, 12:58 AM
$\int \frac{x^3}{\sqrt{1 - 2x^2 }} dx$
$\int \frac{ x . x^2}{u}\frac{\sqrt{1-2x^2}}{2x}du$
$-\frac{1}{2} \int {\frac{\frac{1-u^2}{2}}{u}}du$
$-\frac{1}{2} \int \frac{1-u^2}{2u}du$