A question pertaining to Indefinite Integrals, and Anti-Deriviatives:
My problem: "integral[((9r^2)(dr))/(sqrt.(1-r^3))] , u = (1- r^3)"

**...I know that's ugly... I'm using Safari on an iPad, which goofs up some OS features like inserting pictures, and the equations toolbar thing. Sorry! :(

So, in school, we are evaluating these using "U Substitution", and I am wondering if:
1. The denominator can be written as to the negative half power, and come up.
2. ...what do I do with the "r^2" in "9r^2", if u = (1- r^3)?

Any and all help is greatly appreciated! :)

Question

A question pertaining to Indefinite Integrals, and Anti-Deriviatives: 
My problem: "integral[((9r^2)(dr))/(sqrt.(1-r^3))] , u = (1- r^3)"

**...I know that's ugly... I'm using Safari on an iPad, which goofs up some OS features like inserting pictures, and the equations toolbar thing. Sorry! :(

So, in school, we are evaluating these using "U Substitution", and I am wondering if:
1. The denominator can be written as to the negative half power, and come up.
2. ...what do I do with the "r^2" in "9r^2", if u = (1- r^3)?

Any and all help is greatly appreciated! :)

amonoconnor · Answer

I have a fair understanding of these problems (started them today in Calc), but this one is stumping me. If someone could show the work of them working through the problem, step by step, that would be the most beneficial, and greatly appreciated! (I am a very visual learner, and seeing the process really helps me understand what I'm doing)

ganeshie8 · Answer

$\large \int  \frac{9r^2 dr}{\sqrt{1-r^3}}$

ganeshie8 · Answer

are u able to see latex on ur ipad correctly ?

amonoconnor · Answer

Yes!!

ganeshie8 · Answer

Good :)

U-substitution involves 3 steps

amonoconnor · Answer

...sorry, I just meant that what you posted was the right integral... What's latex?

ganeshie8 · Answer

nvm, i see ipad is processing latex correctly :)

ganeshie8 · Answer

step1 :  make some substitution (in present example, $u = 1-r^3$ )

ganeshie8 · Answer

we're done wid step1 already, right ?

amonoconnor · Answer

Yes, and I understand that much:)

ganeshie8 · Answer

step2 : find $du$ in terms of  $dr$

ganeshie8 · Answer

lets see what it means by working out $du$ in present problem

amonoconnor · Answer

Is it: 
du = ( 0 - 3r^2) dr  ?

ganeshie8 · Answer

Yes !

ganeshie8 · Answer

$\large \int  \frac{9r^2 dr}{\sqrt{1-r^3}}$

substitute $\large  u = 1-r^3$

that gives  $\large   du =  -3r^2 dr$
$\large   \implies  r^2dr =   \frac{-du}{3}   $

ganeshie8 · Answer

plug them in ur integrand

amonoconnor · Answer

...can the denominator be written as (1-r^3)^(-1/2) ?

amonoconnor · Answer

As if multiplied to 9r^2?

ganeshie8 · Answer

you can do that

ganeshie8 · Answer

$\large \int  \frac{9r^2 dr}{\sqrt{1-r^3}}$

substitute $\large  u = 1-r^3$

that gives $\large   du =  -3r^2 dr$
$\large   \implies  r^2dr =   \frac{-du}{3}$

then the integrand becomes :

$\large \int  \frac{9(\frac{-du}{3})}{\sqrt{u}}$

after some simplification :

$\large -3\int  u^{\frac{-1}{2}}du$

ganeshie8 · Answer

^^let me knw if smthng doesnt make sense

amonoconnor · Answer

*working it out on my paper... I'll let you know:)

ganeshie8 · Answer

okie :)

amonoconnor · Answer

...I get what you did, and I follow, but just for clarification : why did you not, when "isolating" dr, divide by -3r^2, and only by -3??

ganeshie8 · Answer

you could do that if u prefer

ganeshie8 · Answer

I didnt divide by entire $-3r^2$, cuz i saw that the numerator has $r^2dr$ term already
so, i thought of replacing $r^2 dr$ with something

amonoconnor · Answer

Ohhh... I see... You worked smart, not hard:)

ganeshie8 · Answer

hahah, next u knw what needs to be done ?

amonoconnor · Answer

Yep, u^(-1/2) = u^1/2 + C so, the final answer is "3u^1/2 + C" ?

amonoconnor · Answer

*plz be right..

ganeshie8 · Answer

nope

amonoconnor · Answer

. . . Rage. . .

ganeshie8 · Answer

$\large  \int x^n dx   =  \frac{x^{n+1}}{n+1} + c$

amonoconnor · Answer

Oh! Do we plug 1-r^3 back in ?

amonoconnor · Answer

Oh, yeah, I forgot about that part...  *more coffee

ganeshie8 · Answer

$\large \int  \frac{9r^2 dr}{\sqrt{1-r^3}}$

substitute $\large  u = 1-r^3$

that gives $\large   du =  -3r^2 dr$
$\large   \implies  r^2dr =   \frac{-du}{3}$

then the integrand becomes :

$\large \int  \frac{9\frac{-du}{3}}{\sqrt{u}}$

after some simplification :

$\large -3\int  u^{\frac{-1}{2}}du $

this integral is easy to evaluate now, evaluating it gives :

$\large -3\frac{u^{\frac{-1}{2}+1}}{\frac{-1}{2}+1} + c $


$\large -6 u^{\frac{1}{2}} + c $


$\large -6 (1-r^3)^{\frac{1}{2}} + c $

ganeshie8 · Answer

yes, you're right!
we need to substitute back the "u" s

ganeshie8 · Answer

"u" is just a passing cloud, it was not present in original integral,
we substituted it in the middle... and so we should replace it by r's in the end

amonoconnor · Answer

Jesus, you're amazing...  :) thank you so much!! You're a lifesaver, once again! I have one more problem on my assignment for tonight, and it looks pretty psycho... Will you be on OS for a while? In case I need your brilliance to guide my understanding again? I want to try it myself, but just in case...