TRIGONOMETRIC SUBSTITUTIONS (i think)

[ineedhelpnow]

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

 

[Opalg]

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

Hi ineedhelpnow, and welcome to MHB!

You have made the correct substitution $x=6\sin\theta$, and I expect that you also substituted $dx = 6\cos\theta\,d\theta$. If so, you should have found yourself looking at the integral of some multiple of $\tan^2\theta$. To continue from there, remember that $\tan^2\theta = \sec^2\theta - 1.$

For help with LaTeX coding, see http://mathhelpboards.com/latex-tips-tutorials-56/mhb-latex-guide-pdf-1142.html. But you have been very careful to use parentheses to make that integral easy to read, even without the LaTeX code! (Clapping)

 

[ineedhelpnow]

yeah that's where I am at. i know this part is simple but I am kinda stuck

 

[Opalg]

yeah that's where I am at. i know this part is simple but I am kinda stuck

So \(\displaystyle \int\tan^2\theta\,d\theta = \int(\sec^2\theta - 1)\,d\theta.\) Can you think of a function (a very familiar one) whose derivative is $\sec^2\theta$ ?

 

[ineedhelpnow]

tan

- - - Updated - - -

oooh ok now i get what to do. super simple...kinda embarrassing that i asked but thanks!

 

[ineedhelpnow]

i got (x/(sqrt(36-x^2)) - sin^-1 (x/6) + C

does that sound right?

 

[Prove It]

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

$\displaystyle \begin{align*} \frac{x}{ \left( 36 - x^2 \right) ^{\frac{3}{2}} } - \int{ \left( 36 - x^2 \right) ^{-\frac{1}{2}} \, \mathrm{d}x } &= \frac{x}{ \left( 36- x^2 \right) ^{\frac{3}{2}}} - \int{ \left\{ 36 - \left[ 6\sin{(\theta)} \right] ^2 \right\} ^{-\frac{1}{2}} \, 6\cos{(\theta)}\,\mathrm{d}\theta } \\ &= \frac{x}{ \left( 36-x^2 \right) ^{\frac{3}{2}}} - 6 \int{ \left\{ 36 \left[ 1 - \sin^2{(\theta)} \right] \right\}^{-\frac{1}{2}} \cos{(\theta)}\,\mathrm{d}\theta } \\ &= \frac{x}{ \left( 36 - x^2 \right) ^{\frac{3}{2}} } - 6\int{ \frac{\cos{(\theta)}}{6\cos{(\theta)}}\,\mathrm{d}\theta } \\ &= \frac{x}{ \left( 36 - x^2 \right) ^{\frac{3}{2}}} - \int{1 \, \mathrm{d}\theta} \\ &= \frac{x}{\left( 36- x^2 \right) ^{\frac{3}{2}}} - \theta + C \\ &= \frac{x}{ \left( 36 - x^2 \right) ^{\frac{3}{2}}} - \arcsin{ \left( \frac{x}{6} \right) } + C \end{align*}$