MHB Steward e6 7r33 rational integral

[karush]

$$\displaystyle
\int\frac{{x}^{2}}{\left(4-{x}^{2}\right)^{3/2}} \ dx$$
$u=4-{x}^{2} \ \ \ du=-2x dx \ \ \ x=\sqrt{4-u}$
$$\displaystyle
\int\frac{4-u}{
\left(u\right)^{3/2}} -2 \sqrt{4-u}\ du$$

Stuck

 

[MarkFL]

Try a trigonometric substitution instead. :)

 

[Prove It]

$$\displaystyle
\int\frac{{x}^{2}}{\left(4-{x}^{2}\right)^{3/2}} \ dx$$
$u=4-{x}^{2} \ \ \ du=-2x dx \ \ \ x=\sqrt{4-u}$
$$\displaystyle
\int\frac{4-u}{
\left(u\right)^{3/2}} -2 \sqrt{4-u}\ du$$

Stuck

Another option is to write $\displaystyle \begin{align*} \int{ \frac{x^2}{\left( 4 - x^2 \right) ^{\frac{3}{2}}}\,\mathrm{d}x } = -\frac{1}{2}\int{ x\,\left( -\frac{2\,x}{\left( 4 - x^2 \right) ^{\frac{3}{2}}} \right) \,\mathrm{d}x } \end{align*}$ and then apply integration by parts with $\displaystyle \begin{align*} u = x \end{align*}$ and $\displaystyle \begin{align*} \mathrm{d}v = -\frac{2\,x}{\left( 4 - x^2 \right) ^{\frac{3}{2}}}\,\mathrm{d}x \end{align*}$.

 

[karush]

$$\displaystyle
I=\int\frac{{x}^{2}}{\left(4-{x}^{2}\right)^{3/2}} \ dx$$
$x=2\sin\left({u}\right)
\ \ \ dx=2\cos\left({u}\right) du
\ \ \ u=\arcsin\left({\frac{x}{2}}\right)$
$$\displaystyle
I=\int\frac{8\sin\left({u}\right)\cos\left({u}\right)}
{\left(4\left(1-\sin^2\left({u}\right)\right)\right)^{3/2}}\ du
\implies\int\tan^2\left({u}\right) \ du
\implies\tan\left({u}\right)-u+C$$
$\sin\left({u}\right)=\frac{x}{2}
\ \ \ \ \tan\left({u}\right)=\frac{x}{\sqrt{4-{x}^{2}}}$
$$I=\frac{x}{\sqrt{4-{x}^{2}}}-\arcsin\left({\frac{x}{2}}\right)+C$$

 

[Prove It]

$$\displaystyle
I=\int\frac{{x}^{2}}{\left(4-{x}^{2}\right)^{3/2}} \ dx$$
$x=2\sin\left({u}\right)
\ \ \ dx=2\cos\left({u}\right) du
\ \ \ u=\arcsin\left({\frac{x}{2}}\right)$
$$\displaystyle
I=\int\frac{8\sin\left({u}\right)\cos\left({u}\right)}
{\left(4\left(1-\sin^2\left({u}\right)\right)\right)^{3/2}}\ du$$

Should be $\displaystyle \begin{align*} \int{ \frac{8\sin^2{(u)}\cos{(u)}}{ \left\{ 4\,\left[ 1 - \sin^2{(u)} \right] \right\} ^{\frac{3}{2}} } \,\mathrm{d}u } \end{align*}$

 

[karush]

$\tiny\text{Stewart e6 {7r33} } $
$$\displaystyle
I=\int\frac{{x}^{2}}{\left(4-{x}^{2}\right)^{3/2}} \ dx$$
$x=2\sin\left({u}\right)
\ \ \ dx=2\cos\left({u}\right) du
\ \ \ u=\arcsin\left({\frac{x}{2}}\right)$
$$\displaystyle
I=\int\frac{8\sin^2\left({u}\right)\cos\left({u}\right)}
{\left(4\left(1-\sin^2\left({u}\right)\right)\right)^{3/2}}\ du
\implies\int\tan^2\left({u}\right) \ du
\implies\tan\left({u}\right)-u+C$$
$\sin\left({u}\right)=\frac{x}{2}
\ \ \ \ \tan\left({u}\right)=\frac{x}{\sqrt{4-{x}^{2}}}$
$$I=\frac{x}{\sqrt{4-{x}^{2}}}-\arcsin\left({\frac{x}{2}}\right)+C$$
$\tiny\text
{from Surf the Nations math study group}$