ineedhelpnow said:
my prof. gave us a bunch of homework questions and i can't seem to get around this one.
integrate (x^2)/((36-x^2)^(3/2) dx (sorry. I am new to this website and i couldn't really understand how to use the commands. don't have much experience with coding. (Speechless)
so i started out using the trig substitution x=6sintheta but i ended totally messing up and i don't know where i went wrong
pleeease help me! thanks
$\displaystyle \begin{align*} \int{\frac{x^2}{\left( 36 - x^2 \right) ^{\frac{3}{2}}}\,\mathrm{d}x } &= -\frac{1}{2}\int{ x \left[ -\frac{2x}{ \left( 36 - x^2 \right) ^{\frac{3}{2}}} \right]\,\mathrm{d}x} \end{align*}$
Now applying integration by parts with $\displaystyle \begin{align*} u = x \implies \mathrm{d}u = \mathrm{d}x \end{align*}$ and $\displaystyle \begin{align*} \mathrm{d}v = -\frac{2x}{\left( 36 - x^2 \right) ^{\frac{3}{2}}} \, \mathrm{d}x \end{align*}$. Making the substitution $\displaystyle \begin{align*} w = 36 - x^2 \implies \mathrm{d}w =-2x \, \mathrm{d}x \end{align*}$ giving $\displaystyle \begin{align*} \mathrm{d}v = w^{-\frac{3}{2}}\,\mathrm{d}w \implies v = \frac{w^{-\frac{1}{2}}}{-\frac{1}{2}} = -\frac{2}{w^{\frac{1}{2}}} = -\frac{2}{ \left( 36 - x^2 \right) ^{\frac{1}{2}}} \end{align*}$ and the integral becomes
$\displaystyle \begin{align*} -\frac{1}{2}\int{ x \left[ -\frac{2x}{\left( 36 - x^2 \right) ^{\frac{3}{2}}} \right] \, \mathrm{d}x } &= -\frac{1}{2} \left[ -\frac{2x}{ \left( 36 - x^2 \right) ^{\frac{3}{2}}} - \int{ -\frac{2}{ \left( 36 - x^2 \right) ^{\frac{1}{2}} } \, \mathrm{d}x} \right] \\ &= \frac{x}{ \left( 36 - x^2 \right) ^{\frac{3}{2}}} - \int{ \left( 36 - x^2 \right) ^{-\frac{1}{2}}\, \mathrm{d}x } \end{align*}$
Now make the substitution $\displaystyle \begin{align*} x = 6\sin{(\theta)} \implies \mathrm{d}x = 6\cos{(\theta)}\,\mathrm{d}\theta \end{align*}$ and the integral becomes
$\displaystyle \begin{align*} \frac{x}{ \left( 36 - x^2 \right) ^{\frac{3}{2}} } - \int{ \left( 36 - x^2 \right) ^{-\frac{1}{2}} \, \mathrm{d}x } &= \frac{x}{ \left( 36- x^2 \right) ^{\frac{3}{2}}} - \int{ \left\{ 36 - \left[ 6\sin{(\theta)} \right] ^2 \right\} ^{-\frac{1}{2}} \, 6\cos{(\theta)}\,\mathrm{d}\theta } \\ &= \frac{x}{ \left( 36-x^2 \right) ^{\frac{3}{2}}} - 6 \int{ \left\{ 36 \left[ 1 - \sin^2{(\theta)} \right] \right\}^{-\frac{1}{2}} \cos{(\theta)}\,\mathrm{d}\theta } \\ &= \frac{x}{ \left( 36 - x^2 \right) ^{\frac{3}{2}} } - 6\int{ \frac{\cos{(\theta)}}{6\cos{(\theta)}}\,\mathrm{d}\theta } \\ &= \frac{x}{ \left( 36 - x^2 \right) ^{\frac{3}{2}}} - \int{1 \, \mathrm{d}\theta} \\ &= \frac{x}{\left( 36- x^2 \right) ^{\frac{3}{2}}} - \theta + C \\ &= \frac{x}{ \left( 36 - x^2 \right) ^{\frac{3}{2}}} - \arcsin{ \left( \frac{x}{6} \right) } + C \end{align*}$
