Compute the integral of 1/(x*sqrt(x^3-25)) Hint: let u=x^(3/2)

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Compute the integral of 1/(x*sqrt(x^3-25)) Hint: let u=x^(3/2)

Mathematics
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I don't know that your hint gets you there. Who gave the hint?
it is on the review sheet
i know the answer too if that helps. its 2/15 arcsec(x^(3/2)/5)+c

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In doing it, they manipulated in a way to use trig method.
Yeah, you have to brush up on your trig method (some variation of tan^2=sec2-1). Using hint x under the square root turns to sq root of u^2-25 and you can do the trig method.
put (x^(3/2)-25)=t^2
use this substitution
I think he wants us to use the trig method. I just keep getting stuck. I have u=x^(3/2) du=3/2sqrt(x)dx 2/3du=sqrt(x)dx I would be able to solve from there but i dont have a sqrt(x)
use my method and multiply denominator and numerator by x^0.5 and put x^(3/2) as t^2+25
He's going to test us on the trig method...
who?
the teacher
oh...but even ur hint uses substitution
right, but isnt there a difference between trig substitution and regular substitution?
if u want to use trig then use sec^(2/3)t=x
got it?
not really. Can you go through the process step by step?
use (25*sec (t))^(4/3)=x so x^(3/2)=25*sec^2(t) so dx=25^(4/3)*(4/3)*sec^(1/3)(t)*sec(t)*tan(t)
dx=c*sec^(4/3)(t)*tan(t) dt c=constant
now solve further i m very slow in typing
Jas, give me a minute. I have the method his instructor is expecting.
OK, using hint \[u=x ^{3/2}\]Convert this all the way through you get \[u ^{2/3}=x\]
if u will solve my expression then it will become simplified
Yes, but he is in school, he is learning a really helpful method, technique.
Just hold on one minute, let me finish, so the work don't get all muddled.
i have already told similar method before but he wants trig method
k......go ahead
\[2/3 du =\sqrt{x}\]
if u will use x^(3/2)-25=t^2 then it will be more easier
New\[\int\limits_{?}^{?}[(2/3) du]/[(u ^{2/3}\sqrt{u ^{2}-25}\]
what happened to 2/3du=sqrt(x)?
it is 2/3du=sqrt(x)dx
\[\int\limits_{?}^{?}[(2/3) du]/25\sec \theta \tan \theta\]
I guess that should be derivative theta
i mean, what happens to the sqrt(x)?
multiply deno and nume by x and cha... made mistake in calculation
In the substitution there is an extra \[\sqrt{5 \sec \theta}\] and that is = to sq rt of x. That is swallowed up in dtheta.
OK, I just noticed the square root of x is in bottom. How do I get it on top to be swallowed by dtheta, Jas?
you solve complete expression in u first it will be 1/u*(......)
(.....)=u^2-25
now put u=sec(t)
One second, there is a fix for this, don't change it too much or uswag would get lost in translation.
otherwise change directly which i already proposed earlier
OK, that is already wiped out uswag. when I bring sq rt x and dx on top, I have an extra sq rt x in bottom, they cancel each other out.
there is no sqrt but there is x
Yes, but that x is converted to u^(2/3). In my conversion I keep u^2=5 sec theta and I wipe out sq root 5 theta or sq rt x.
Uswag, you seem to be following along. Does this make any sense?
k...i was doing in other way
i need to go
Can we start the explanation from 2/3 du/ u^(2/3)(sqrt(u^2-25)..... i got lost in the explanation after that. so its 2/3 du= sqrt(x)dx so we have sqrt(x)dx on top.
Thanks Jas, you been a great help.
Thanks Jas
thanks 4 being my fan
\[{[2/(3\sqrt{x})]du}/[\sqrt{x}u ^{2}\sqrt{u ^{2}-25}]\]
\[2/75\int\limits_{?}^{?}d \theta/\sec \theta \tan \theta\]
Now you have a function you can integrate. Once you integrate you have to go back and change to x. These trig techniques are explained nicely in MIT lectures http://www.youtube.com/watch?v=CXKoCMVqM9s&feature=BFa&list=PL590CCC2BC5AF3BC1&index=25
I'll write down some of the substitutions I made\[u^{2}-25=5^{2}\sec ^{2}\theta-5^{2}\]
And that is equal to \[5^{2}\tan ^{2}\theta\]
\[u =5\sec \theta \]
\[\sqrt{u ^{2}-25}=\sqrt{5^{2}\sec ^{2}\theta-5^{2}}=\sqrt{5^{2}\tan ^{2}\theta}=5\tan \theta\]
A lot of info, makes sense?
yeah, after seeing the info plus that it does. Thanks!

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