Int. by Substitution: int [ (9r^2) / (sqrt[1 - r^3]) ] dr

[kaebun]

I'm not realy sure how to do this, I have a vague idea, but I think I'm skipping steps, and I'm not sure I got the right answer. If someone could show me the proper way , that would be realy helpful.

the problem: \(\displaystyle \int((9r^2 dr)/(sqrt(1-r^3))\) u=1-r^3

my work:

\(\displaystyle du= -3r^2 dr\)
\(\displaystyle \int((9r^2dr)/sqt(u))=\)
\(\displaystyle 2sqt(u) ((9r^2 dr)/ (-3r^2)) +c=\)

\(\displaystyle -6 sqrt(1-r^3) +c\)

Thanks, Kathy

 

[galactus]

\(\displaystyle \L\\9\int\frac{r^{2}}{\sqrt{1-r^{3}}}dr\)

Let \(\displaystyle \L\\u=\sqrt{1-r^{3}}, \;\ du=\frac{-3r^{2}}{2\sqrt{1-r^{3}}}dx, \;\ \frac{-2}{3}du=\frac{r^{2}}{\sqrt{1-r^{3}}}\)

 

[kaebun]

I'm not sure I understand. You want the side with du to equal the function you are taking a derivitive of?
If thats true then, in the problem it says that u=1- r^3 then du=-3x^2 which isn't even close to working properly

 

[kaebun]

I looked up some examples on the internet, and found this great step by step thing , it makes so much more sense now. But I do have one more question, although I know how to get the right answer, I don't realy understand why it works. Could someone explain that?

 

[galactus]

The substituions I outlined work as well as any.

See what \(\displaystyle \frac{-2}{3}du\) is equal to?.

Make the substitutions and all you have to integrate is:

\(\displaystyle \L\\9\int\frac{-2}{3}du\)

Which gives:

\(\displaystyle \L\\-6u\)

Resub:

\(\displaystyle \L\\-6\sqrt{1-x^{3}}\)....Voila!

For a check differentiate and see if you get back to the original.

\(\displaystyle \L\\\frac{d}{dr}[-6\sqrt{1-r^{3}}]\)

Product rule:

\(\displaystyle {-6}(\frac{-3r^{2}}{2\sqrt{1-r^{3}}})+\sqrt{1-r^{3}}(0)\)

\(\displaystyle \L\\\frac{18r^{2}}{2\sqrt{1-r^{3}}}\)

\(\displaystyle \L\\\frac{9r^{2}}{\sqrt{1-r^{3}}}\)

There we go. Checks.

 

[kaebun]

I know that other substitutions would work, but in the book we were told to use that specific one. I've figured it out now,
THanks for your help