anonymous
  • anonymous
Compute the integral of 1/(x*sqrt(x^3-25)) Hint: let u=x^(3/2)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
I don't know that your hint gets you there. Who gave the hint?
anonymous
  • anonymous
it is on the review sheet
anonymous
  • anonymous
i know the answer too if that helps. its 2/15 arcsec(x^(3/2)/5)+c

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anonymous
  • anonymous
In doing it, they manipulated in a way to use trig method.
anonymous
  • anonymous
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.
anonymous
  • anonymous
put (x^(3/2)-25)=t^2
anonymous
  • anonymous
use this substitution
anonymous
  • anonymous
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)
anonymous
  • anonymous
use my method and multiply denominator and numerator by x^0.5 and put x^(3/2) as t^2+25
anonymous
  • anonymous
He's going to test us on the trig method...
anonymous
  • anonymous
who?
anonymous
  • anonymous
the teacher
anonymous
  • anonymous
oh...but even ur hint uses substitution
anonymous
  • anonymous
right, but isnt there a difference between trig substitution and regular substitution?
anonymous
  • anonymous
if u want to use trig then use sec^(2/3)t=x
anonymous
  • anonymous
got it?
anonymous
  • anonymous
not really. Can you go through the process step by step?
anonymous
  • anonymous
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)
anonymous
  • anonymous
dx=c*sec^(4/3)(t)*tan(t) dt c=constant
anonymous
  • anonymous
now solve further i m very slow in typing
anonymous
  • anonymous
Jas, give me a minute. I have the method his instructor is expecting.
anonymous
  • anonymous
OK, using hint \[u=x ^{3/2}\]Convert this all the way through you get \[u ^{2/3}=x\]
anonymous
  • anonymous
if u will solve my expression then it will become simplified
anonymous
  • anonymous
Yes, but he is in school, he is learning a really helpful method, technique.
anonymous
  • anonymous
Just hold on one minute, let me finish, so the work don't get all muddled.
anonymous
  • anonymous
i have already told similar method before but he wants trig method
anonymous
  • anonymous
k......go ahead
anonymous
  • anonymous
\[2/3 du =\sqrt{x}\]
anonymous
  • anonymous
if u will use x^(3/2)-25=t^2 then it will be more easier
anonymous
  • anonymous
New\[\int\limits_{?}^{?}[(2/3) du]/[(u ^{2/3}\sqrt{u ^{2}-25}\]
anonymous
  • anonymous
what happened to 2/3du=sqrt(x)?
anonymous
  • anonymous
it is 2/3du=sqrt(x)dx
anonymous
  • anonymous
\[\int\limits_{?}^{?}[(2/3) du]/25\sec \theta \tan \theta\]
anonymous
  • anonymous
I guess that should be derivative theta
anonymous
  • anonymous
i mean, what happens to the sqrt(x)?
anonymous
  • anonymous
multiply deno and nume by x and cha... made mistake in calculation
anonymous
  • anonymous
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.
anonymous
  • anonymous
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?
anonymous
  • anonymous
you solve complete expression in u first it will be 1/u*(......)
anonymous
  • anonymous
(.....)=u^2-25
anonymous
  • anonymous
now put u=sec(t)
anonymous
  • anonymous
One second, there is a fix for this, don't change it too much or uswag would get lost in translation.
anonymous
  • anonymous
otherwise change directly which i already proposed earlier
anonymous
  • anonymous
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.
anonymous
  • anonymous
there is no sqrt but there is x
anonymous
  • anonymous
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.
anonymous
  • anonymous
Uswag, you seem to be following along. Does this make any sense?
anonymous
  • anonymous
k...i was doing in other way
anonymous
  • anonymous
i need to go
anonymous
  • anonymous
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.
anonymous
  • anonymous
Thanks Jas, you been a great help.
anonymous
  • anonymous
Thanks Jas
anonymous
  • anonymous
thanks 4 being my fan
anonymous
  • anonymous
\[{[2/(3\sqrt{x})]du}/[\sqrt{x}u ^{2}\sqrt{u ^{2}-25}]\]
anonymous
  • anonymous
\[2/75\int\limits_{?}^{?}d \theta/\sec \theta \tan \theta\]
anonymous
  • anonymous
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
anonymous
  • anonymous
I'll write down some of the substitutions I made\[u^{2}-25=5^{2}\sec ^{2}\theta-5^{2}\]
anonymous
  • anonymous
And that is equal to \[5^{2}\tan ^{2}\theta\]
anonymous
  • anonymous
\[u =5\sec \theta \]
anonymous
  • anonymous
\[\sqrt{u ^{2}-25}=\sqrt{5^{2}\sec ^{2}\theta-5^{2}}=\sqrt{5^{2}\tan ^{2}\theta}=5\tan \theta\]
anonymous
  • anonymous
A lot of info, makes sense?
anonymous
  • anonymous
yeah, after seeing the info plus that it does. Thanks!

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